3.6: Capítulo 3 Ejercicios con soluciones

3.6: Capítulo 3 Ejercicios con soluciones

3.1 Ejercicios

 
 

Ejercicio ( PageIndex {1} )

 

Screen Shot 2019-09-06 at 8.06.34 PM.png

 

Jodiah está ahorrando su dinero para comprar un sistema de juegos Playstation 3. Estima que necesitará $ 950 para comprar la unidad en sí, los accesorios y algunos juegos. Tiene ahorrados $ 600 en este momento, y razonablemente puede poner $ 60 en sus ahorros al final de cada mes. Dado que la cantidad de dinero ahorrada depende de cuántos meses hayan pasado, elija el tiempo, en meses, como su variable independiente y colóquelo en el eje horizontal. Deje que t represente el número de meses transcurridos y marque cada mes. Elija el dinero ahorrado, en dólares, como su variable dependiente y colóquelo en el eje vertical. Deje que A represente la cantidad ahorrada en dólares. Como Jodiah ahorra $ 60 cada mes, será conveniente dejar que cada cuadro represente $ 60. Copie el siguiente sistema de coordenadas en una hoja de papel cuadriculado.

 

Screen Shot 2019-09-06 at 8.04.26 PM.png

 

a) En el mes 0, Jodiah tiene $ 600 ahorrados. Esto corresponde al punto (0, 600). Trace este punto en su sistema de coordenadas.

 

b) Durante el mes siguiente, ahorró $ 60 más. Comenzando en el punto (0, 600), muévase 1 mes a la derecha y $ 60 hacia arriba y trace un nuevo punto de datos. ¿Cuáles son las coordenadas de este punto?

 

c) Cada vez que vaya a la derecha 1 mes, debe subir $ 60 y trazar un nuevo punto de datos. Repita este proceso hasta llegar al borde del sistema de coordenadas.

 

d) Teniendo en cuenta que estamos modelando esta situación discreta continuamente, dibuje una línea a través de sus puntos de datos.

 

e) Usa tu gráfica para estimar cuánto dinero habrá ahorrado Jodiah después de 7 meses. f) Usando su gráfica, calcule cuántos meses le llevará haber ahorrado suficiente dinero para comprar su sistema de juego, accesorios y juegos.

 
     
Respuesta
     
     

d)

     

屏幕快照 2019-09-17 下午2.33.36.png

     

e)

     

屏幕快照 2019-09-17 下午2.34.55.png

     

En el gráfico, cuando t = 7 meses, habrá ahorrado $ 1020.

     

f)

     

屏幕快照 2019-09-17 下午2.36.01.png

     

Tenga en cuenta que hemos modelado un problema discreto continuamente: ahorra $ 60 al final de cada mes, y tendrá $ 900 al final del quinto mes; y luego $ 960 para fines del mes seis. No habrá tiempo en el que tenga exactamente $ 950, por lo que la respuesta es de 6 meses, momento en el que tendrá $ 960.

     
 
 
 
 

Ejercicio ( PageIndex {2} )

 

Screen Shot 2019-09-06 at 8.07.10 PM.png

 

El letrero de arriba muestra los precios de un viaje en taxi de Liberty Cab Company. Como el costo depende de la distancia recorrida, haga que la distancia sea la variable independiente y colóquela en el eje horizontal. Sea d la distancia recorrida, en millas. Debido a que la compañía de taxis cobra por cada 1/6 de milla, es conveniente marcar cada 1/6 de milla. Haga que el precio, en $, sea su variable dependiente y colóquelo en el eje vertical. Deje que C represente el costo, en $. Debido a que el costo ocurre en incrementos de 40c, marque cada 40c a lo largo del eje vertical. Copie el siguiente sistema de coordenadas en una hoja de papel cuadriculado.

 

Screen Shot 2019-09-06 at 8.09.12 PM.png

 

a) Para los primeros 1/6 de milla de viaje, el costo es de $ 2.30. Esto corresponde al punto (1/6, $ 2.30). Trace este punto en su sistema de coordenadas.

 

b) Para el próximo 1/6 de milla, el costo aumenta en 40 ( cent ). Comenzando en el punto (1/6, $ 2.30), muévase 1/6 de milla a la derecha y 40c hacia arriba y trace un nuevo punto de datos. ¿Cuáles son las coordenadas de este punto?

 

c) Cada vez que vaya a la derecha 1/6 de milla, debe subir 40 ( cent ) y trazar un nuevo punto de datos. Repita este proceso hasta llegar al borde de su sistema de coordenadas.

 

d) Teniendo en cuenta que estamos modelando esta situación discreta continuamente, dibuje una línea a través de sus puntos de datos.

 

e) Melissa se sube a un taxi en la ciudad de Niagara Falls, a unas 2 millas del Parque Estatal Niagara Falls. Use su gráfica para estimar la tarifa del parque.

 

f) En otra parte de la zona, Georgina toma un taxi. Ella solo tiene $ 5 por la tarifa. Use el gráfico para estimar qué tan lejos puede viajar, en millas, con solo $ 5 por la tarifa.

 
 
 

Ejercicio ( PageIndex {3} )

 

Un barco está a 200 pies de una boya en el mar. Se acerca a la boya a una velocidad promedio de 15 pies / s.

 

a) Eligiendo el tiempo, en segundos, como su variable independiente y la distancia desde la boya, en pies, como su variable dependiente, haga un gráfico de un sistema de coordenadas en una hoja de papel cuadriculado que muestre los ejes y las unidades. Use marcas de verificación para identificar sus escalas.

 

b) En el tiempo t = 0, el bote está a 200 pies de la boya. ¿A qué punto corresponde esto? Trace este punto en su sistema de coordenadas.

 

c) Después de 1 segundo, el bote se ha acercado 15 pies más cerca de la boya. Comenzando en el punto anterior, muévase 1 segundo hacia la derecha y 15 pies hacia abajo (ya que la distancia está disminuyendo) y trace un nuevo punto de datos. ¿Cuáles son las coordenadas de este punto?

 

d) Cada vez que vaya a la derecha 1 segundo, debe bajar 15 pies y trazar un nuevo punto de datos. Repita este proceso hasta llegar a 12 segundos.

 

e) Dibuje una línea a través de sus puntos de datos.

 

f) Cuando el bote está a 50 pies de la boya, el conductor quiere comenzar a reducir la velocidad. Use su gráfica para estimar qué tan pronto el bote estará a 50 pies de la boya.

 
     
Respuesta
     
     

e)

     

屏幕快照 2019-09-17 下午2.37.15.png

     

f) Dibujamos una línea a 50 pies y vemos que ocurre a los 10 segundos:

     

屏幕快照 2019-09-17 下午2.38.08.png

     
 
 
 
 

Ejercicio ( PageIndex {4} )

 

Joe debe $ 24,000 en préstamos estudiantiles. Ha terminado la universidad y ahora está trabajando. Puede permitirse pagar $ 1500 por mes para sus préstamos.

 

a) Elija el tiempo en meses como su variable independiente y el monto adeudado, en $, como la variable dependiente. En una hoja de papel cuadriculado, haga un bosquejo del sistema de coordenadas, usando marcas de verificación y etiquetando los ejes apropiadamente.

 

b) En el tiempo t = 0, Joe aún no ha pagado nada por sus préstamos. ¿A qué punto corresponde esto? Trace este punto en su sistema de coordenadas.

 

c) Después de un mes, paga $ 1500. Comenzando en el punto anterior, mueva 1 mes a la derecha y $ 1500 hacia abajo (hacia abajo porque la deuda está disminuyendo). Traza este punto. ¿Cuáles son sus coordenadas?

 

d) Cada vez que vaya 1 mes a la derecha, debe mover $ 1500 hacia abajo. Continúe haciendo esto hasta que sus préstamos hayan sido pagados.

 

e) Teniendo en cuenta que estamos modelando esta situación discreta continuamente, dibuje una línea a través de sus puntos de datos.

 

f) Use el gráfico para determinar cuántos meses le tomará pagar el monto total de sus préstamos.

 
 
 

Ejercicio ( PageIndex {5} )

 

Screen Shot 2019-09-06 at 8.13.53 PM.png

 

Earl, la ardilla solo tiene diez días más para la hibernación. Necesita salvar 50 bellotas más. Está cansado de recolectar bellotas, por lo que solo puede recolectar 8 bellotas cada 2 días.

 

a) Deje que t represente el tiempo en días y conviértalo en su variable independiente. Deje que N represente el número de bellotas recolectadas y conviértalo en su variable dependiente. Configure un sistema de coordenadas a escala apropiada en una hoja de papel cuadriculado.

 

b) En el tiempo t = 0, Earl ha recogido cero de las bellotas que necesita. ¿A qué punto corresponde esto? Trace este punto en su sistema de coordenadas.

 

c) Después de dos días (t = 2), Earl ha recogido 8 bellotas. Comenzando en el punto anterior, muévase 2 días hacia la derecha y 8 bellotas hacia arriba. Traza este punto. ¿Cuáles son sus coordenadas?

 

d) Cada vez que vaya 2 días a la derecha, debe mover 8 bellotas hacia arriba y trazar un punto. Continúa haciendo esto hasta que alcances los 14 días.

 

e) Teniendo en cuenta que estamos modelando esta situación discreta continuamente, dibuje una línea a través de sus puntos de datos.

 

f) Usa el gráfico para determinar cuántas bellotas habrá recogido después de 10 días. ¿Earl habrá recolectado suficientes bellotas para su hibernación de invierno?

 

g) Observe que el número de bellotas recolectadas aumenta a razón de 8 bellotas cada 2 días. Reduzca esto a una tasa que indique la cantidad promedio de bellotas que se recolecta cada día.

 

h) La siguiente tabla enumera la cantidad de bellotas que Earl habrá recolectado en varias ocasiones. Algunas de las entradas se han completado para usted. Por ejemplo, en t = 0, Earl no tiene bellotas, entonces N = 0. Después de un día, la cantidad aumenta en 4, entonces N = 0 + 4 (1). Después de dos días, se han producido dos aumentos, por lo que N = 0 + 4 (2). El patrón continúa. Completa las entradas que faltan.

                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                           
t N
0 0
1 0 + 4 (1)
2 0 + 4 (2)
3 0 + 4 (3)
4
6
8
10
12
14
 

i) Exprese el número de bellotas recolectadas, N, en función del tiempo t, en días.

 

j) Use su función para predecir la cantidad de bellotas que Earl tendrá después de 10 días. ¿Esta respuesta está de acuerdo con su estimación de la parte (f)?

 
     
Respuesta
     
     

e)

     

屏幕快照 2019-09-17 下午2.39.19.png

     

f) Si dibujas una línea a los 10 días, entonces puedes ver que habrá recogido 40 bellotas.

     

屏幕快照 2019-09-17 下午2.40.08.png

     

g) ( frac {8} {2} ) bellotas / día = 4 bellotas / día

     

h) Siguiendo el patrón, obtenemos:

                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                               
t N
0 0
1 0 + 4 (1)
2 0 + 4 (2)
3 0 + 4 (3)
4 0 + 4 (4)
6 0 + 4 (6)
8 0 + 4 (8)
10 0 + 4 (10)
12 0 + 4 (12)
14 0 + 4 (14)
     
 
 

i) N = 0 + 4t o N = 4t

 

j) En t = 10, N = 0 + 4 (10) = 40 bellotas.

 
 
 

Ejercicio ( PageIndex {6} )

 

En la televisión en red, una hora típica de programación contiene 15 minutos de comerciales y anuncios y 45 minutos del programa en sí.

 

a) Elija la cantidad de televisión vista como su variable independiente y colóquela en el eje horizontal. Deje que T represente la cantidad de televisión que se ve, en horas. Elija la cantidad total de comerciales / anuncios vistos como su variable dependiente y colóquelos en el eje vertical. Deje que C represente la cantidad total de comerciales / anuncios vistos, en minutos. Usando una hoja de papel cuadriculado, haga un boceto de un sistema de coordenadas y etiquete adecuadamente.

 

b) Durante 0 horas de programación observada, se han visto 0 minutos de comerciales. ¿A qué punto corresponde esto? Grafícalo en tu sistema de coordenadas.

 

c) Después de ver 1 hora de programación, se han visto 15 minutos de comerciales / anuncios. Comenzando en el punto anterior, muévase 1 hora hacia la derecha y 15 minutos hacia arriba. Traza este punto. ¿Cuáles son sus coordenadas?

 

d) Cada vez que vaya 1 hora a la derecha, debe moverse 15 minutos hacia arriba y trazar un punto. Continúe haciendo esto hasta que alcance las 5 horas de programación.

 

e) Dibuje una línea a través de sus puntos de datos.

 

f) Billy ve la televisión durante cinco horas el lunes. Use el gráfico para determinar cuántos minutos de comerciales ha visto durante este tiempo.

 

g) Suponga que una persona ha visto una hora de comerciales / anuncios. Usa el gráfico para estimar cuántas horas de televisión vio.

 

h) La siguiente tabla muestra el número de horas de programación observadas en relación con el número de minutos de anuncios / anuncios vistos. Durante 0 horas de televisión, se ven 0 minutos de comerciales / anuncios. Por cada hora de televisión que veamos, debemos contar 15 minutos de comerciales / anuncios. Entonces, durante 1 hora, se miran 0 + 15 (1) minutos de comerciales. Durante 2 horas, 0 + 15 (2) minutos; y así. Completa las entradas que faltan.

                                                                                                                                                                                                                                                                                                                                                                           
T (horas) C (minutos)
0 0
1 0 + 15 (1)
2 0 + 15 (2)
3
4
5
 

i) Exprese la cantidad de comerciales / anuncios vistos, C, en función de la cantidad de televisión que ve T. Use su ecuación para predecir la cantidad de anuncios / anuncios vistos durante 5 horas de programación televisiva. ¿Esta respuesta está de acuerdo con su estimación de la parte (f)?

 
 
 

Ejercicio ( PageIndex {7} )

 

Según la OTAN (Asociación Nacional de Propietarios de Teatros), el precio promedio de un boleto de cine fue de 5,65 dólares en el año 2001. Desde entonces, el precio promedio ha aumentado cada año en aproximadamente 20 ( cents ) .

 

a) Elija el año, comenzando con 2000, como la variable independiente y haga marcas cada año en el eje. Elija el precio promedio del boleto, en dólares, como su variable dependiente y comience en 5.65 dólares, con marcas cada 10 ( cents ) ​​arriba. Haga un boceto de un sistema de coordenadas y etiquete adecuadamente.

 

b) En 2001, el precio promedio de las entradas fue de 5,65 dólares, correspondiente al punto (2001, 5,65). Grafícalo en tu sistema de coordenadas.

 

c) En 2002, un año después, el precio promedio aumentó en aproximadamente 20 ( cents ). Comenzando en el punto anterior, muévase a la derecha por 1 año y hacia arriba por 20 ( cents ) ​​y trace el punto. ¿Cuáles son sus coordenadas?

 

d) Cada vez que vaya 1 año a la derecha, debe avanzar 20 ( cents ) ​​y trazar un punto. Continúe haciendo esto hasta el año 2010.

 

e) Teniendo en cuenta que estamos modelando esta situación discreta continuamente, dibuje una línea a través de sus puntos de datos.

 

f) Use la gráfica para estimar en qué año el precio promedio de un boleto pasará 7.00 dólares.

 
     
Respuesta
     
     

e)

     

屏幕快照 2019-09-17 下午2.45.33.png

     

f) Dibuje una línea por $ 7.00 y busque el año.

     

屏幕快照 2019-09-17 下午2.46.31.png

     

Observe que el año es entre 2007 y 2008. Pero este es un problema discreto, ya que solo estamos tratando con años enteros. Por lo tanto, la respuesta es 2008.

     
 
 
 
 

Ejercicio ( PageIndex {8} )

 

Cuando Jessica conduce su automóvil a una conferencia relacionada con el trabajo, su empleador le reembolsa aproximadamente 45 centavos por milla para cubrir el costo del combustible y el desgaste del vehículo.

 

a) Usando la distancia recorrida d, en millas, como la variable independiente y la cantidad reembolsada A, en dólares, como la variable dependiente, haga un bosquejo de un sistema de coordenadas y etiquete apropiadamente. Marque la distancia cada 5 millas y la cantidad reembolsada cada $ 0.45.

 

b) Para viajar 0 millas, el reembolso es 0. Esto corresponde al punto (0, 0). Grafícalo en tu sistema de coordenadas.

 

c) Para un viaje que requiere que conduzca un total de 5 millas, se le reembolsa 5 (0.45) = $ 2.25. Esto corresponde al punto (5, $ 2.25). Tramalo.

 

d) Por cada 5 millas que vaya hacia la derecha, debe subir $ 2.25 y trazar el punto. Haga esto hasta llegar a 20 millas.

 

e) Teniendo en cuenta que estamos modelando esta situación discreta continuamente, dibuje una línea a través de sus puntos de datos.

 

f) En marzo, Jessica asiste a una conferencia que está a solo 5 millas de distancia. Contando ida y vuelta, recorre 10 millas en total. Use la gráfica para determinar cuánto le reembolsan.

 

g) En diciembre, ella asiste a una conferencia a 10 millas de distancia. ¿Cuánto dura su viaje en total? Use la gráfica para determinar cuánto le reembolsarán.

 

h) Para viajes más largos, como 200 millas totales, probablemente necesitarás hacer un gráfico mucho más grande. ¿Y si viaja 400 millas? O más? Son limitaciones como estas las que hacen que sea útil encontrar una ecuación que describa lo que muestra el gráfico. Para encontrar la ecuación, comenzamos con una tabla que nos ayuda a comprender la relación entre las variables dependientes e independientes. Completa la tabla de abajo.

                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                           
d (millas) A ($)
0 0
1 0 + 0,45 (1)
2 0 + 0,45 (2)
3
4
5
10
20
50
100
 

i) Use la tabla de la parte (h) para obtener una ecuación que relacione dy A.

 

j) Ahora, use la ecuación para determinar los montos de reembolso para viajes de 200 millas y 400 millas.

 
 
 

Ejercicio ( PageIndex {9} )

 

La temperatura se mide típicamente en grados Fahrenheit en los Estados Unidos; pero se mide en grados Celsius en muchos otros países. La relación entre Fahrenheit y Celsius es lineal. Elija la medida de grados en grados Celsius para que sea nuestra variable independiente y la medida de grados en grados Fahrenheit para que sea nuestra variable dependiente. El agua se congela a 0 grados Celsius, que corresponde a 32 grados Fahrenheit; y el agua hierve a 100 grados centígrados, lo que corresponde a 212 grados Fahrenheit. Podemos trazar esta información como los dos puntos (0,32) y (100,212). La relación es lineal, así que tenga el siguiente gráfico:

 

Screen Shot 2019-09-06 at 8.37.00 PM.png

 

a) Use el gráfico para aproximar la temperatura equivalente de Fahrenheit para 48 grados Celsius.

 

b) Para determinar la tasa de cambio de Fahrenheit con respecto a Celsius, dibujamos un triángulo rectángulo con lados paralelos a los ejes que conecta los dos puntos que conocemos …

 

Screen Shot 2019-09-06 at 8.38.57 PM.png

 

El lado PR tiene una longitud de 100 grados, lo que representa un aumento de 100 grados centígrados. Side RQ es 180 grados, lo que representa un aumento de 180 grados Fahrenheit. Encuentra la tasa de aumento de Fahrenheit por Celsius. c) La siguiente tabla muestra algunos valores de temperaturas en grados Celsius y sus correspondientes lecturas de Fahrenheit. Cero grados Celsius corresponde a 32 grados Fahrenheit. Nuestra tasa es de 9 grados Fahrenheit por cada 5 grados Celsius, o 9/5 de grado Fahrenheit por cada 1 grado Celsius. Entonces, para 1 grado Celsius, aumentamos la lectura de Fahrenheit en 9/5 grados, obteniendo 32 + 9/5 (1). Para 2 grados Celsius, aumentamos en dos casos de 9/5 grados para obtener 32 + 9/5 (2). Complete las entradas que faltan, siguiendo el patrón.

                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                           
C (grados) F (grados)
0 32
1 (32 + frac {9} {5} (1) )
2 (32 + frac {9} {5} (2) )
3 (32 + frac {9} {5} (3) )
4
5
10
20
48
100
 

d) Usa la tabla para formar una ecuación que dé grados Fahrenheit en términos de grados Celsius.

 
     
Respuesta
     
     

a) Hacemos una línea a 48 grados Celsius y leemos la estimación de Fahrenheit.

     

屏幕快照 2019-09-17 下午2.49.21.png

     

La estimación debe ser de aproximadamente 120 grados Fahrenheit.

     

b) ( dfrac {changeinF} {changeinC} = dfrac {180} {100} = dfrac {9} {5} )

     

c)

                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                               
C (grados) F (grados)
0 32
1 (32 + frac {9} {5} (1) )
2 (32 + frac {9} {5} (2) )
3 (32 + frac {9} {5} (3) )
4 (32 + frac {9} {5} (4) )
5 (32 + frac {9} {5} (5) )
10 (32 + frac {9} {5} (10) )
20 (32 + frac {9} {5} (20) )
48 (32 + frac {9} {5} (48) )
100 (32 + frac {9} {5} (100) )
     
 
 

d) F = 9 5C + 32

 
 
 

Ejercicio ( PageIndex {10} )

 

El 16 de junio de 2006, la tasa de conversión de euros a dólares estadounidenses era de aproximadamente 0,8 a 1, lo que significa que cada 0,8 euros valía 1 dólar estadounidense.

 

a) Eligiendo dólares para ser la variable independiente y euros para ser la variable dependiente, haga una gráfica del sistema de coordenadas. Marque cada dólar en el eje del dólar y cada 0,8 euros en el eje del euro. Etiquetar adecuadamente.

 

b) Cero dólares valen 0 euros. Esto corresponde al punto (0, 0). Grafícalo en tu sistema de coordenadas.

 

c) Un dólar vale 0,8 euros. Trace esto como un punto en su sistema de coordenadas.

 

d) Por cada dólar que mueva hacia la derecha, debe subir 0,8 euros y trazar un punto. Haga esto hasta llegar a $ 10.

 

e) Dibuje una línea a través de sus puntos de datos.

 

f) Usa la gráfica para estimar cuántos euros valen $ 8.

 

g) Usa la gráfica para estimar cuántos dólares valen 5 euros.

 

h) La siguiente tabla muestra algunos valores de dólares y su valor correspondiente en euros. Completa las entradas que faltan.

                                                                                                                                                                                                                                                                                                                                                                                                                           
Dólares Euros
0 0
1 0 + 0,8 (1)
2 0 + 0,8 (2)
3
4
5
10
 

i) Use la tabla para hacer una ecuación que se pueda usar para convertir dólares a euros.

 

j) Use la ecuación de (i) para convertir $ 8 a euros. ¿Su respuesta está de acuerdo con la respuesta de (f) que obtuvo usando la gráfica?

 
 
 

Ejercicio ( PageIndex {11} )

 

La Torre de Pisa en Italia tiene su famosa inclinación hacia el sur porque el suelo de arcilla y arena en el que está construido es más suave en el lado sur que en el norte. La inclinación a menudo se encuentra midiendo la distancia que la parte superior de la torre sobresale de la base, indicada por h en la figura a continuación. En 1980, la torre tenía una inclinación de h = 4.49m, y esta inclinación aumentaba en aproximadamente 1 mm / año.

 

Screen Shot 2019-09-06 at 8.54.42 PM.png

 

Figura ( PageIndex {1} ). h mide la inclinación de la Torre de Pisa.

 

Investigaremos cómo cambió la inclinación de la torre de 1980 a 1995.

 

a) Primero, tenga en cuenta que nuestras unidades no coinciden: la inclinación en 1980 fue de 4,49 m, pero el aumento anual de la inclinación es de 1 mm / año. Nuestro primer objetivo es hacer que las unidades sean iguales. Usaremos milímetros (mm). Convertir 4,49 ma mm.

 

b) Obtenga una hoja de papel cuadriculado. Dado que la inclinación de la torre depende del año, convierta el año en la variable independiente y colóquelo en el eje horizontal. Deje que t represente el año. Convierta la inclinación en la variable dependiente y colóquela en el eje vertical. Sea h la inclinación, medida en milímetros (mm). Elija 1980 como el primer año en el eje horizontal y marque cada año a partir de entonces, hasta 1995. Deje que el eje vertical comience a 4,49 m, convertido a mm desde la parte (a), ya que esa fue nuestra primera medición; y luego marcamos cada 1 mm hasta 4510 mm.

 

c) Piense en 1980 como el año de inicio. Junto con la medición de inclinación de ese año, forma un punto. ¿Cuáles son las coordenadas de este punto? Trace el punto en su sistema de coordenadas.

 

d) Comenzando en el primer punto, desde la parte (c), muévase un año a la derecha (hasta 1981) y 1 mm hacia arriba (porque la inclinación aumenta) y trace un nuevo punto de datos.

 

e) Cada vez que se mueve un año a la derecha, debe moverse 1 mm hacia arriba y trazar un nuevo punto. Repita este proceso hasta llegar al año 1995.

 

f) Teniendo en cuenta que estamos modelando esta situación discreta continuamente, dibuje una línea a través de sus puntos de datos. Podemos usar este modelo para hacer predicciones.

 

g) De acuerdo con los modelos de simulación por computadora, que utilizan matemáticas sofisticadas, la torre estaría en peligro de colapsar cuando h alcance aproximadamente 4495 mm. Usa tu gráfica para estimar en qué año ocurrirá esto.

 

h) En realidad, la inclinación de la torre pasó de 4495 mm y la torre no colapsó. De hecho, la inclinación aumentó a 4500 mm antes de que la torre se cerrara el 7 de enero de 1990, para someterse a renovaciones para disminuir la inclinación. (La torre se volvió a abrir en 2001, después de que los ingenieros usaran pesas y eliminaran la suciedad debajo de la base para disminuir la inclinación en 450 mm.) ¿Cuáles podrían ser algunas de las razones por las cuales la predicción del modelo de la computadora era incorrecta?

 

i) La siguiente tabla enumera la inclinación de la torre, h, el año y la cantidad de años transcurridos desde 1980. En 1980, la inclinación era de 4490 mm y todavía no había ocurrido el aumento de 1 mm, por lo que complete 4490 + 0 (1) = 4490. En 1981, se produjo una ocurrencia del aumento de 1 mm porque había pasado un año desde 1980. Por lo tanto, la inclinación fue de 4490 + 1 (1). En 1982, se produjeron dos casos del aumento de 1 mm, ya que habían pasado 2 años desde 1980. Por lo tanto, la inclinación fue de 4490 + 2 (1). Y el patrón continúa de esta manera. Rellene las entradas restantes.

                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                         
Año años x después de ’80 inclinación h
1980 0 4490 + 0 (1)
1981 1 4490 + 1 (1)
1982 2 4490 + 2 (1)
1983
1984
1985
1986
1987
1988
1989
1990
1991
1992
1993
1994
1995
 

j) Sea x el número de años transcurridos desde 1980 y h represente la inclinación. Usando la tabla de arriba, escribe una ecuación que relacione h y x.

 

k) Usa tu ecuación para predecir la inclinación en 1990. ¿Está de acuerdo con el valor real de 1990? ¿Está de acuerdo con el valor que se muestra en el gráfico que hizo?

 

l) En la parte (g), usaste el gráfico para predecir el año en que la inclinación sería de 4495 mm. Usa tu ecuación para hacer la misma predicción. ¿Están de acuerdo las respuestas?

 
     
Respuesta
     
     

a) Hay 1000 mm en 1 m, entonces 4.49 = 4.49 (1000) = 4490 mm.

     

f)

     

屏幕快照 2019-09-17 下午2.53.51.png

     

g) Dibujamos una línea para h = 4495 y vemos que corresponde a 1985.

     

屏幕快照 2019-09-17 下午2.55.19.png

     

h) Ningún modelo es perfecto. El modelo de computadora no debe haber tenido en cuenta ciertos factores inesperados.

     

i)

                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                 
Año años x después de ’80 inclinación h
1980 0 4490 + 0 (1)
1981 1 4490 + 1 (1)
1982 2 4490 + 2 (2)
1983 3 4490 + 2 (3)
1984 4 4490 + 2 (4)
1985 5 4490 + 2 (5)
1986 6 4490 + 2 (6)
1987 7 4490 + 2 (7)
1988 8 4490 + 2 (8)
1989 9 4490 + 2 (9)
1990 10 4490 + 2 (10)
1991 11 4490 + 2 (11)
1992 12 4490 + 2 (12)
1993 13 4490 + 2 (13)
1994 14 4490 + 2 (14)
1995 15 4490 + 2 (15)
     
 
 

j) h = 4490 + 1x

 

k) En 1990, x = 10, y entonces h = 4490 + 1 (10) = 4500 mm. Sí, concuerda con el valor real en 1990.

 

l) Para determinar cuándo la inclinación será 4495, establezca h = 4495 y resuelva para x. 4495 = 4490 + 1x conduce a 5 = x, por lo que nuestra respuesta es 1985. Esto concuerda con la respuesta de (g).

 
 
 

Ejercicio ( PageIndex {12} )

 

De acuerdo con el Resumen estadístico de los Estados Unidos (www.census.gov), hubo aproximadamente 31,000 crímenes reportados en los Estados Unidos en 1998, y esto disminuyó a una tasa de aproximadamente 2900 por año.

 

a) En una hoja de papel cuadriculado, haga un sistema de coordenadas y trace los datos de 1998 como un punto. Note that you will only need to graph the first quadrant of a coordinate system, since there are no data for years before 1998 and there cannot be a negative number of crimes reported. Use the given rate to find points for 1999 through 2006, and then draw a line through your data. We are constructing a continuous model for our discrete situation.

 

b) The following table lists the number of crimes reported, C, the year, and the number of years since 1998. In 1998, the number was 31, 000 and no occurrences of the 2900 decrease had happened yet, so we fill in 31000 − 2900(0). In 1999, one occurrence of the 2900 decrease had happened because one year had passed since 1998. Therefore, the number of crimes reported was 31000−2900(1). And the pattern continues in this manner. Fill in the remaining entries.

                                                                                                                                                                                                                                                                                                                                                                                                  
Year yrs x after 1998 No. of crimes C
1998 0 31000 − 2900(0)
1999 1 31000 − 2900(1)
2000    
2001    
2002    
 

c) Observing the pattern in the table, we come up with the equation C = 31000 − 2900x to relate the number of crimes C to the number of years x after 1998. Here, C is a function of x, and so we can use the notation C(x) = 31000 − 2900x to emphasize this.

 

i. Compute C(5).

 

ii. In a complete sentence, explain what C(5) represents.

 

iii. Compute C(8).

 

iv. In a complete sentence, explain what C(8) represents.

 
 
 

Exercise (PageIndex{13})

 

According to the Statistical Abstract of the United States (www.census.gov), there were approximately 606, 000 inmates in United States prisons in 1999, and this was increasing by a rate of about 14, 000 per year.

 

a) On a sheet of graph paper, make a coordinate system and plot the 1999 data as a point. Note that you will only need to graph the first quadrant of a coordinate system, since there are no data for years before 1999 and there cannot be a negative number of crimes reported. Use the given rate to find points for 2000 through 2006, and then draw a line through your data. We are constructing a continuous model for our discrete situation.

 

b) The following table lists the number of inmates, N, the year, and the number of years since 1999. In 1999, the number was 606, 000 and no occurrences of the 14, 000 increase had happened yet, so we fill in 606000 + 14000(0). In 2000, one occurrence of the 14, 000 increase had happened because one year had passed since 1999. Therefore, the number of crimes reported was 606000 + 14000(1). And the pattern continues in this manner. Fill in the remaining entries.

                                                                                                                                                                                                                                                                                                                                                 
Year yrs x after ’99 No. of inmates N
1999 0 606000+14000(0)
2000 1 606000+14000(1)
2001    
2002    
 

c) Observing the pattern in the table, we come up with the equation N = 606000+14000x to relate the number of crimes C to the number of years x after 1999. Here, N is a function of x, and so we can use the notation N(x) = 606000+14000x to emphasis this.

 

i. Compute N(5).

 

ii. In a complete sentence, explain what N(5) represents.

 

iii. Compute N(7).

 

iv. In a complete sentence, explain what N(7) represents.

 
     
Answer
     
     

a)
屏幕快照 2019-09-17 下午3.00.24.png

     

b)

                                                                                                                                                                                                                                                                                                                                                                                                                                                         
Year yrs x after ’99 No. of inmates N
1999 0 606000+14000(0)
2000 1 606000+14000(1)
2001 2 606000+14000(2)
2002 3 606000+14000(3)
     
 
 

 c)

 

i. N(5) = 606000 + 14000(5) = 676000.

 

ii. It means that, according to our model, 5 years after 1999 (that is, in 2004), the number of inmates will be 676, 000.

 

iii. N(7) = 606000 + 14000(7) = 704000.

 

iv. It means that, according to our model, in 2006, the number of inmates will be 704, 000

 
 

3.2 Exercises

 
 

Exercise (PageIndex{1})

 

Suppose you are riding a bicycle up a hill as shown below.

 

屏幕快照 2019-09-17 下午3.03.40.png

 

Figure (PageIndex{1}). Riding a bicycle up a hill.

 

a) If the hill is straight as shown, consider the slant, or steepness, of its incline. As you ride up the hill, what can you say about the slant? Does it change? If so, how?

 

b) The slant is what mathematicians call the slope. To confirm your answer to part (a), you will place the hill on a coordinate system and compute its slope along various segments of the hill. See the figure below.

 

屏幕快照 2019-09-17 下午3.05.12.png

 

Three points–P, Q and R–have been labeled along the hill. We call the vertical distance (height) the rise and the horizontal distance the run. As you ride up the hill from point P to point Q, what is the rise? What is the run? Use these values to compute the slope from P to Q.

 

c) Now consider as you ride from P to R. What is the rise? What is the run? Use these values to compute the slope from P to R.

 

d) Finally, consider as you ride from Q to R. What is the rise? What is the run? Use these values to compute the slope from Q to R.

 

e) How do the values for slope from parts (b)-(d) compare? Do these results confirm your answer to part (a)?

 

f) Notice that the slope is positive in this example. In this context of riding a bicycle over a hill, what would negative slope mean?

 
     
Answer
     
     

a) No, it does not change. The slant is the same everywhere along the straight hill.

     

b) (m_{PQ} = dfrac{3−1 }{9−3} = dfrac{2}{6} = dfrac{1}{3}) 

     

c)  (m_{PR} = dfrac{4−1 }{12−3} = dfrac{3}{9} = dfrac{1}{3}) 

     

d)  (m_{QR} = dfrac{4-3}{12−9} = dfrac{1}{3}) 

     

e) They are all the same. This makes sense because the slant or steepness of the hill is the same throughout.

     

f) Positive slope means that you are riding uphill; negative slope would mean that you are riding downhill.

     
 
 
 
 

Exercise (PageIndex{2})

 

Set up a coordinate system on a sheet of graph paper, plotting the points P(3, 4) and Q(−2, −7) and drawing the line through them.

 

a) What can you say about the slope of the line? Is it positive, zero, negative or undefined? Is the slope the same everywhere along the line, or does it change in places? If it does change, where are the slopes different?

 

b) Use your graph to determine the change in y (rise) and the change in x (run). Use these results to compute the slope of the line.

 

c) Use the slope formula to compute the slope of the line.

 

d) Does your numerical solution from part (c) agree with your graphical solution from part (b)? If not, check your work for errors.

 
 
 

Exercise (PageIndex{3})

 

Set up a coordinate system on a sheet of graph paper, plotting the points P(−1, 3) and Q(5, −3) and drawing the line through them.

 

a) What can you say about the slope of the line? Is it positive, zero, negative or undefined? Is the slope the same everywhere along the line, or does it change in places? If it does change, where are the slopes different?

 

b) Use your graph to determine the change in y (rise) and the change in x (run). Use these results to compute the slope of the line.

 

c) Use the slope formula to compute the slope of the line.

 

d) Does your numerical solution from part (c) agree with your graphical solution from part (b)? If not, check your work for errors.

 
     
Answer
     
     

a) The slope is negative because the line slants downhill. The slope is the same everywhere along the line because the slant of the line does not change.

     

b)

     

 

     

  屏幕快照 2019-09-17 下午3.40.00.png

     

slope = −6/6 = −1

     

c) ∆y = −3 − (3) = −6; ∆x = 5 − (−1) = 6; slope = dfrac{delta y} {delta x}) =( dfrac{−6 }{6}) = −1

     

d) Yes.

     
 
 
 

In Exercises (PageIndex{4})-(PageIndex{10}), perform each of the following tasks.

 

i. Make a sketch of a coordinate system; plot the given points, and draw the line through the points.

 

ii. Use the slope formula to compute the slope of the line through the given points. Reduce the slope where possible.

 
 

Exercise (PageIndex{4})

 

(0, 0) and (3, 4)

 
 
 

Exercise (PageIndex{5})

 

(−5, 2) and (0, 3)

 
     
Answer
     
     

屏幕快照 2019-09-17 下午3.45.13.png

     

(slope = dfrac{3−2}{ 0−(−5)} = dfrac{1}{5})

     
 
 
 
 

Exercise (PageIndex{6})

 

(−3, −3) and (6, −5)

 
 
 

Exercise (PageIndex{7})

 

(2, 0) and (2, 2)

 
     
Answer
     
     

屏幕快照 2019-09-17 下午3.47.18.png

     

(slope = dfrac{2−0 }{2−2} = dfrac{2}{0}) = undefined

     
 
 
 
 

Exercise (PageIndex{8})

 

(−9, −3) and (6, −3)

 
 
 

Exercise (PageIndex{9})

 

(−8, 4) and (3, −8)

 
     
Answer
     
     

屏幕快照 2019-09-17 下午3.48.39.png

     

(slope = dfrac{−8−4}{ 3−(−8)} = dfrac{−12}{ 11})

     
 
 
 
 

Exercise (PageIndex{10})

 

(−2, 6) and (5, −2)

 
 
 

Exercise (PageIndex{11})

 

For the following line, two convenient points P and Q have been chosen. We chose two points that were at the corners of boxes on our grid so their coordinates are easy to read.

 

屏幕快照 2019-09-17 下午3.12.48.png

 

a) Label their coordinates.

 

b) Thinking of P as the starting point and Q as the ending point, draw a right triangle joining the points.

 

c) Clearly state the change in y (rise) and the change in x (run) from P to Q.

 

d) Compute the slope.

 
     
Answer
     
     

a) The points are (0, 0) and (6, 3).

     

b)

     

屏幕快照 2019-09-17 下午3.52.08.png

     

c) ∆y = 3 − 0 = 3; ∆x = 6 − 0 = 6

     

d) slope = (dfrac{delta y}{delta x} =dfrac{3}{6} = dfrac{1}{ 2})

     
 
 
 
 

Exercise (PageIndex{12})

 

For the following line, two convenient points A and B have been chosen. We chose two points that were at the corners of boxes on our grid so their coordinates are easy to read.

 

屏幕快照 2019-09-17 下午3.15.52.png

 

a) Label their coordinates.

 

b) Thinking of A as the starting point and B as the ending point, draw a right triangle joining the points.

 

c) Clearly state the change in y (rise) and the change in x (run) from A to B.

 

d) Compute the slope.

 
 
 

Exercise (PageIndex{13})

 

Copy the coordinate system below onto a sheet of graph paper. Then do the following:

 

a) Select any two convenient points P and Q on the graph of the line. Label each point with its coordinates.

 

b) Clearly state the change in y (rise) and the change in x (run). Compute the slope of the line.

 

屏幕快照 2019-09-17 下午3.17.41.png

 
     
Answer
     
     

NOTE: Solutions may vary depending on which two convenient points were chosen.

     

a) You can pick any two points on the line; for example, (0, 0) and (5, 4) as shown below.

     

屏幕快照 2019-09-17 下午3.57.14.png

     

b) ∆y = 4 − 0 = 4; ∆x = 5 − 0 = 5; slope = (dfrac{delta y }{delta x} = dfrac{4}{5})

     
 
 
 
 

Exercise (PageIndex{14})

 

Copy the coordinate system below onto a sheet of graph paper. Then do the following:

 

a) Select any two convenient points P and Q on the graph of the line. Label each point with its coordinates.

 

b) Clearly state the change in y (rise) and the change in x (run). Compute the slope of the line.

 

屏幕快照 2019-09-17 下午3.18.58.png

 
 
 

Exercise (PageIndex{15})

 

Copy the coordinate system below onto a sheet of graph paper. Then do the following:

 

a) Select any two convenient points P and Q on the graph of the line. Label each point with its coordinates.

 

b) Clearly state the change in y (rise) and the change in x (run). Compute the slope of the line.

 

屏幕快照 2019-09-17 下午3.20.18.png

 
     
Answer
     
     

NOTE: Solutions may vary depending on which two convenient points were chosen.

     

a) You can pick any two points on the line; for example, (1, 1) and (3, 7) as shown below.

     

屏幕快照 2019-09-17 下午3.59.24.png

     

b) ∆y = 7 − 1 = 6; ∆x = 3 − 1 = 2; slope =( dfrac{delta y }{delta x} = dfrac{6}{2} = 3)

     
 
 
 
 

Exercise (PageIndex{16})

 

Copy the coordinate system below onto a sheet of graph paper. Then do the following:

 

a) Select any two convenient points P and Q on the graph of the line. Label each point with its coordinates.

 

b) Clearly state the change in y (rise) and the change in x (run). Compute the slope of the line.

 

屏幕快照 2019-09-17 下午3.21.51.png

 
 
 

Exercise (PageIndex{17})

 

The following coordinate system shows the graphs of three lines, each with different slope. Match each slope with (a), (b), or (c) appropriately.

 

slope = 1

 

slope = 2/3

 

slope = −2

 

屏幕快照 2019-09-17 下午3.23.04.png

 
     
Answer
     
     

slope = 1: (b)

     

slope = 2/3: (c)

     

slope = −2: (a)

     
 
 
 
 

Exercise (PageIndex{18})

 

The following coordinate system shows the graphs of three lines, each with different slope. Match each slope with (a), (b), or (c) appropriately.

 

slope = 2

 

slope = −1/3

 

slope = 1/2

 

屏幕快照 2019-09-17 下午3.25.20.png

 
 
 

Exercise (PageIndex{19})

 

Draw a coordinate system on a sheet of graph paper for which the x- and y-axes both range from −10 to 10.

 

a) Draw a line that contains the point (0, 1) and has slope 2. Label the line as (a).

 

b) On the same coordinate system, draw a line that contains the point (0, 1) and has slope −1/2. Label it as (b).

 

c) Use the slopes of these two lines to show that they are perpendicular.

 
     
Answer
     
     

b)

     

屏幕快照 2019-09-17 下午4.04.23.png

     

c) (m_{1}m_{2} = 2(−1/2) = −1), so the lines are perpendicular.

     
 
 
 
 

Exercise (PageIndex{20})

 

Draw a coordinate system on a sheet of graph paper for which the x- and y-axes both range from −10 to 10.

 

a) Draw a line that contains the point (1, −2) and has slope 1/3. Label the line as (a).

 

b) On the same coordinate system, draw a line that contains the point (0, 1) and has slope −3. Label it as (b).

 

c) Use the slopes of these two lines to show that they are perpendicular.

 
 
 

Exercise (PageIndex{21})

 

Draw a line through the point P(1, 3) that is parallel to the line through the origin with slope −1/4.

 
     
Answer
     
     

屏幕快照 2019-09-17 下午4.05.44.png

     
 
 
 
 

Exercise (PageIndex{22})

 

Draw a line through the point P(1,3) that is parallel to the line through the origin with slope 3/5.

 
 
 

Exercise (PageIndex{23})

 

Draw a coordinate system on a sheet of graph paper for which the x- and y-axes both range from −10 to 10.

 

a) Draw a line that contains the point (−1, −2) and has slope 3/4. Label the line as (a).

 

b) On the same coordinate system, draw a line that contains the point (0, 1) and has slope 4/3. Label it as (b).

 

c) Are these lines parallel, perpendicular or neither? Show using their slopes.

 
     
Answer
     
     

b)

     

屏幕快照 2019-09-17 下午4.08.14.png

     

c) (m_{1}m_{2} = (4/3)(3/4) = 1 neq −1), so the lines are not perpendicular; the slopes are not equal, so the lines are not parallel, either. Thus, the lines may be classified as neither parallel nor perpendicular.

     
 
 
 
 

Exercise (PageIndex{24})

 

Graph a coordinate system on a sheet of graph paper for which the x- and y-axes both range from −10 to 10.

 

a) Draw a line that contains the point (−4, 0) and has slope 1. Label the line as (a).

 

b) On the same coordinate system, draw a line that contains the point (0, 2) and has slope −1. Label it as (b).

 

c) Are these lines parallel, perpendicular or neither? Show using their slopes.

 
 
 

Exercise (PageIndex{25})

 

屏幕快照 2019-09-17 下午3.33.03.png

 

Figure (PageIndex{2}). A grade is a way of expressing slope.

 

On the road from Fort Bragg to Willits or from Fort Bragg to Santa Rosa, one often passes signs like that shown above. A grade is just slope expressed as a percent instead of a fraction or decimal. In other words, the grade measures the steepness of the road just as slope does.

 

a) An 80 /0 grade means that, for every horizontal distance of 100 ft, the road rises or drops 8 ft (depending on whether you are going uphill or downhill). Write 80 /0 grade as slope in reduced fractional form.

 

b) Suppose a hill drops 16 ft for every 180 ft horizontally. Find the grade of the hill to the nearest tenth of a percent.

 

c) Explain in a complete sentence or sentences what a grade of 00 /0 would represent.

 
     
Answer
     
     

a) grade (=dfrac{ 8}{ 100} = dfrac{2}{ 25})

     

b) grade (= dfrac{16}{ 180} = dfrac{4}{ 45} = 8.90 )%

     

c) 0% grade represents no grade or slope; that is, a flat road.

     
 
 
 

3.3 Exercises 

 

In Exercises (PageIndex{1})-(PageIndex{6}), perform each of the following tasks for the given linear function.

 

i. Set up a coordinate system on a sheet of graph paper. Label and scale each axis. Remember to draw all lines with a ruler.

 

ii. Identify the slope and y-intercept of the graph of the given linear function.

 

iii. Use the slope and y-intercept to draw the graph of the given linear function on your coordinate system. Label the y-intercept with its coordinate and the graph with its equation.

 
 

Exercise (PageIndex{1})

 

f(x) = 2x + 1

 
     
Answer
     
     

Compare f(x) = 2x + 1 with f(x) = mx + b. Note that the slope is m = 2 and the y-coordinate of the y-intercept is b = 1. Therefore, the y-intercept will be the point (0, 1). Plot the point P(0, 1). To obtain a line of slope m = 2/1, start at the point P(0, 1), then move 1 unit to the right and 2 units upward, arriving at the point Q(1, 3), as shown in the figure below. The line through the points P and Q is the required line.

     

屏幕快照 2019-09-21 上午12.46.53.png

     
 
 
 
 

Exercise (PageIndex{2})

 

f(x) = −2x + 3

 
 
 

Exercise (PageIndex{3})

 

f(x) = 3 − x

 
     
Answer
     
     

Compare f(x) = 3 − x, or equivalently f(x) = −x + 3, with f(x) = mx + b. Note that the slope is m = −1 and the y-coordinate of the y-intercept is b = 3. Therefore, the y-intercept will be the point (0, 3). Plot the point P(0, 3). To obtain a line of slope m = −1, start at the point P(0, 3), then move 1 unit to the right and 1 units downward, arriving at the point Q(1, 2), as shown in the figure below. The line through the points P and Q is the required line.

     

屏幕快照 2019-09-21 上午12.48.26.png

     
 
 
 
 

Exercise (PageIndex{4})

 

f(x) = 2 − 3x

 
 
 

Exercise (PageIndex{5})

 

(f(x) = −frac{3}{4} x + 3)

 
     
Answer
     
     

Compare f(x) = (−3/4)x+3 with f(x) = mx+b. Note that the slope is m = −3/4 and the y-coordinate of the y-intercept is b = 3. Therefore, the y-intercept will be the point (0, 3). Plot the point P(0, 3). To obtain a line of slope m = −3/4, start at the point P(0, 3), then move 4 units to the right and 3 units downward, arriving at the point Q(4, 0), as shown in the figure below. The line through the points P and Q is the required line.

     

屏幕快照 2019-09-21 上午12.49.32.png

     
 
 
 
 

Exercise (PageIndex{6})

 

(f(x) = frac{2}{3} x − 2)

 
 

In Exercises (PageIndex{7})-(PageIndex{12}), perform each of the following tasks.

 

i. Make a copy of the given graph on a sheet of graph paper.

 

ii. Label the y-intercept with its coordinates, then draw a right triangle and label the sides to help identify the slope.

 

iii. Label the line with its equation.

 
 

Exercise (PageIndex{7})

 

屏幕快照 2019-09-21 上午12.08.06.png

 
     
Answer
     
     

The slope is found by dividing the rise by the run (see figure). Hence, the slope is 1/2. The y-intercept is found by noting where the graph of the line crosses the y-axis (see figure), in this case, at (0, −3). Hence, m = 1/2 and b = −3, so the equation of the line in slope intercept form is

     

[y = mx + b quad text{or} quad y = frac{1}{2}x − 3]

     

屏幕快照 2019-09-21 上午12.51.44.png

     
 
 
 
 

Exercise (PageIndex{8})

 

屏幕快照 2019-09-21 上午12.08.50.png

 
 
 

Exercise (PageIndex{9})

 

屏幕快照 2019-09-21 上午12.09.30.png

 
     
Answer
     
     

The slope is found by dividing the rise by the run (see figure). Hence, the slope is 2/3. The y-intercept is found by noting where the graph of the line crosses the y-axis (see figure), in this case, at (0, −2). Hence, m = 2/3 and b = −2, so the equation of the line in slope intercept form is

     

[y = mx + b quad text{or} quad y = frac{2}{3}x − 2]

     

屏幕快照 2019-09-21 上午12.53.05.png

     
 
 
 
 

Exercise (PageIndex{10})

 

屏幕快照 2019-09-21 上午12.10.55.png

 
 
 

Exercise (PageIndex{11})

 

屏幕快照 2019-09-21 上午12.11.52.png

 
     
Answer
     
     

The slope is found by dividing the rise by the run (see figure). Hence, the slope is 3/2. The y-intercept is found by noting where the graph of the line crosses the y-axis (see figure), in this case, at (0, 1). Hence, m = 3/2 and b = 1, so the equation of the line in slope intercept form is

     

[y = mx + b quad text{or} quad y = frac{3}{2}x + 1]

     

屏幕快照 2019-09-21 上午12.55.42.png

     
 
 
 
 

Exercise (PageIndex{12})

 

屏幕快照 2019-09-21 上午12.12.29.png

 
 

 

 
 

Exercise (PageIndex{13})

 

Kate makes $39, 000 per year and gets a raise of $1000 each year. Since her salary depends on the year, let time t represent the year, with t = 0 being the present year, and place it along the horizontal axis. Let salary S, in thousands of dollars, be the dependent variable and place it along the vertical axis. We will assume that the rate of increase of $1000 per year is constant, so we can model this situation with a linear function.

 

a) On a sheet of graph paper, make a graph to model this situation, going as far as t = 10 years.

 

b) What is the S-intercept?

 

c) What is the slope?

 

d) Suppose we want to predict Kate’s salary in 20 years or 30 years. We cannot use the graphical model because it only shows up to t = 10 years. We could draw a larger graph, but what if we then wanted to predict 50 years into the future? The point is that a graphical model is limited to what it shows. A model algebraic function, however, can be used to predict for any year! Find the slope-intercept form of the linear function that models Kate’s salary.

 

e) Write the function using function notation, which emphasizes that S is a function of t.

 

f) Now use the algebraic model from (e) to predict Kate’s salary 10 years, 20 years, 30 years, and 50 years into the future.

 

g) Compute S(40).

 

h) In a complete sentence, explain what the value of S(40) from part (g) means in the context of the problem.

 
     
Answer
     
     

a)

     

屏幕快照 2019-09-21 上午12.56.57.png

     

b) At t = 0 (present year), her salary is $39, 000. Since S is in thousands of dollars, S = 39 when t = 0. So the S-intercept is (0, 39).

     

c) The increase in Kate’s salary is $1, 000 per year, but S is in thousands of dollars, so the rate of increase in S is 1. That is, the slope is 1.

     

d) Using the slope-intercept form, we get S = t + 39.

     

e) S(t) = t + 39.

     

f)

     
             
  • To find Kate’s salary in 10 years, compute S(10) = 10 + 39 = 49, which means that she will be earning $49, 000 per year.
  •          
  • To find Kate’s salary in 20 years, compute S(20) = 20 + 39 = 59, which means that she will be earning $59, 000 per year.
  •          
  • To find Kate’s salary in 30 years, compute S(30) = 30 + 39 = 69, which means that she will be earning $69, 000 per year.
  •          
  • To find Kate’s salary in 50 years, compute S(50) = 50 + 39 = 89, which means that she will be earning $89, 000 per year.
  •      
     

g) S(40) = 40 + 39 = 79. h) If the current rate of increase continues, in 40 years Kate’s salary will be $79, 000.

     
 
 
 
 

Exercise (PageIndex{14})

 

For each DVD that Blue Charles Co. sells, they make 5c profit. Profit depends on the number of DVD’s sold, so let number sold n be the independent variable and profit P, in $, be the dependent variable.

 

a) On a sheet of graph paper, make a graph to model this situation, going as far as n = 15.

 

b) Use the graph to predict the profit if n = 10 DVD’s are sold.

 

c) The graphical model is limited to predicting for values of n on your graph. Any larger value of n necessitates a larger graph, or a different kind of model. To begin finding an algebraic model, identify the P-intercept of the graph.

 

d) What is the slope of the line in your graphical model?

 

e) Find a slope-intercept form of a linear function that models Blue Charles Co.’s sales.

 

f) Write the function using function notation.

 

g) Explain why this model does not have the same limitation as the graphical model.

 

h) Find P(100), P(1000), and P(10000).

 

i) In complete sentences, explain what the values of P(100), P(1000), and P(10000) mean in the context of the problem.

 
 
 

Exercise (PageIndex{15})

 

Enrique had $1, 000 saved when he began to put away an additional $25 each month.

 

a) Let t represent time, in months, and S represent Enrique’s savings, in $. Identify which should be the independent and dependent variables.

 

b) To begin finding a linear function to model this situation, identify the S-intercept and slope.

 

c) Find a slope-intercept form of a linear function to model Enrique’s savings over time.

 

d) Write the linear function in function notation.

 

e) Use the function model to predict how much will be in his savings in one year.

 

f) Use the function model to predict when will he have $2000 saved.

 

g) Graph the function on a coordinate system.

 

h) At the same time, Anne-Marie also begins to save $25 per month, but she begins with $1200 already in her savings. Make a graphical model of her situation and place it on the same coordinate system as the graphical model for Enrique’s savings. Label it appropriately.

 

i) How do the lines compare to each other? Say something about their slopes.

 

j) Find a slope-intercept form of a linear function that models Anne-Marie’s savings. Use the same variables as you did for Enrique’s model.

 

k) Write the function using function notation.

 

l) Prove that the graphs of the two functions are parallel lines.

 

m) For Anne-Marie, looking at the graphs, do you think it will take her more time or less time than Enrique to save up $2000?

 

n) Use the linear function model for Anne-Marie to predict how long it will take her to save $2000. Does this agree with your expectation from (m)?

 
     
Answer
     
     

a) t should be the independent variable and S should be the dependent variable.

     

b) S-intercept = (0, 1000); slope = 25

     

c) S = 25t + 1000

     

d) S(t) = 25t + 1000

     

e) S(12) = 25(12) + 1000 = 1300

     

f) Set S=2000 and solve for t. 2000 = 25t + 1000 1000 = 25t 40 = t So it will take 40 months for him to reach $2000.

     

h)

     

屏幕快照 2019-09-21 上午1.04.45.png

     

i) The lines have the same slope; they are parallel.

     

j) S = 25t + 1200

     

k) S(t) = 25t + 1200

     

l) They are lines because they are in the y = mx + b form. They are parallel because their slopes are equal (both are 25).

     

m) It should take her less time because her graph is above Enrique’s graph. This makes sense intuitively since she began with more money than he did.

     

n) Set S=2000 and solve for t. 2000 = 25t + 1200 800 = 25t 32 = t So it will take 32 months for her to reach $2000. This agrees with our expectation from (m): It takes her less time than Enrique

     
 
 
 
 

Exercise (PageIndex{16})

 

Jose is initially 400 meters away from the bus stop. He starts running toward the stop at a rate of 5 meters per second.

 

a) Express Jose’s distance d from the bus stop as a function of time t.

 

b) Use your model to determine Jose’s distance from the bus stop after one minute.

 

c) Use your model to determine the time it will take Jose to reach the bus stop.

 
 
 

Exercise (PageIndex{17})

 

A ball is dropped from rest above the surface of the earth. As it falls, its speed increases at a constant rate of 32 feet per second per second.

 

a) Express the speed v of the ball as a function of time t.

 

b) Use your model to determine the speed of the ball after 5 seconds.

 

c) Use your model to determine the time it will take for the ball to achieve a speed of 256 feet per second.

 
     
Answer
     
     

a) We will do a “rough” plot of speed v versus time t. Speed depends upon time, so we place the speed on the vertical axis and the time on the horizontal axis in the figure that follows. The intial speed is 0 ft/s, which give us the v-intercept at P(0, 0). The rate at which the speed is increasing (acceleration) is constant and will be the slope of the line; i.e., the slope of the line is m = 32 ft/s2 (32 feet per second per second).

     

屏幕快照 2019-09-21 上午1.17.31.png

     

Because we know the slope and intercept of the line, we can use the slope intercept form y = mx + b and substitute m = 32 and b = 0 to obtain

     

[y = mx + b\ y = 32x + 0 \y = 32x]

     

However, we are using v and t in place of y and x, so we replace these in the last formula to obtain

     

[v = 32t]

     

or using function notation,

     

[v(t) = 32t]
b) To find the speed of the ball after 5 seconds, substitute t = 5 into the equation developed in the previous part.

     

[v(t) = 32t \ v(5) = 32(5) \ v(5) = 160]

     

Hence, the speed of the ball after 5 seconds is 160 feet per second.

     

c) To find the time it takes the ball to reach 256 feet per second, we must find t so that v(t) = 256.

     

[v(t) = 256\ 32t = 256\ t = 8]

     

Thus, it takes 8 seconds for the ball to attain a speed of 256 feet per second.

     
 
 
 
 

Exercise (PageIndex{18})

 

A ball is thrown into the air with an initial speed of 200 meters per second. It immediately begins to lose speed at a rate of 9.5 meters per second per second.

 

a) Express the speed v of the ball as a function of time t.

 

b) Use your model to determine the speed of the ball after 5 seconds.

 

c) Use your model to determine the time it will take for the ball to achieve its maximum height.

 
 

In Exercises (PageIndex{19})-(PageIndex{24}), a linear function is given in standard form Ax + By = C. In each case, solve the given equation for y, placing the equation in slope-intercept form. Use the slope and intercept to draw the graph of the equation on a sheet of graph paper.

 
 

Exercise (PageIndex{19})

 

3x − 2y = 6

 
     
Answer
     
     

Place 3x − 2y = 6 in slope-intercept form. First subtract 3x from both sides of the equation, then divide both sides of the resulting equation by −2.

     

[begin{array} {lll} 3x − 2y & = &6 −2 \ y &=& −3x + 6 \ y &= &frac{3}{2}x − 3 end{array}]

     

Compare y = (3/2)x − 3 with y = mx + b to see that the slope is m = 3/2 and the y-coordinate of the y-intercept is b = −3. Therefore, the y-intercept will be the point (0, −3). Plot the point P(0, −3). To obtain a line of slope m = 3/2, start at the point P(0, −3), then move 3 units up and 2 units to the right, arriving at the point Q(2, 0), as shown in the figure below. The line through the points P and Q is the required line.

     

屏幕快照 2019-09-21 上午1.23.13.png

     

 

     
 
 
 
 

Exercise (PageIndex{20})

 

3x + 5y = 15

 
 
 

Exercise (PageIndex{21})

 

3x + 2y = 6

 
     
Answer
     
     

Place 3x + 2y = 6 in slope-intercept form. First subtract 3x from both sides of the equation, then divide both sides of the resulting equation by 2.

     

[begin{array}{lll} 3x + 2y & =& 6 \ 2y &=& −3x + 6 \ y &=& −frac{3}{2} x + 3 end{array}]

     

Compare y = (−3/2)x + 3 with y = mx + b to see that the slope is m = −3/2 and the y-coordinate of the y-intercept is b = 3. Therefore, the y-intercept will be the point (0, 3). Plot the point P(0, 3). To obtain a line of slope m = −3/2, start at the point P(0, 3), then move 3 units downward and 2 units to the right, arriving at the point Q(2, 0), as shown in the figure below. The line through the points P and Q is the required line.

     

屏幕快照 2019-09-21 上午1.25.33.png

     
 
 
 
 

Exercise (PageIndex{22})

 

4x − y = 4

 
 
 

Exercise (PageIndex{23})

 

x − 3y = −3

 
     
Answer
     
     

Place x − 3y = −3 in slope-intercept form. First subtract x from both sides of the equation, then divide both sides of the resulting equation by −3.

     

[begin{array} {lll} x − 3y &=& −3 \  −3y &=& −x − 3 \ y &=& frac{1}{3} x + 1 end{array}]

     

Compare y = (1/3)x + 1 with y = mx + b to see that the slope is m = 1/3 and the y-coordinate of the y-intercept is b = 1. Therefore, the y-intercept will be the point (0, 1). Plot the point P(0, 1). To obtain a line of slope m = 1/3, start at the point P(0, 1), then move 1 unit upward and 3 units to the right, arriving at the point Q(3, 2), as shown in the figure below. The line through the points P and Q is the required line.

     

屏幕快照 2019-09-21 上午1.28.57.png

     
 
 
 
 

Exercise (PageIndex{24})

 

x + 4y = −4

 
 

In Exercises (PageIndex{25})-(PageIndex{30}), you are given a linear function in slope-intercept form. Place the linear function in standard form Ax+ By = C, where A, B, and C are integers and A > 0.

 
 

Exercise (PageIndex{25})

 

(y = frac{2}{3}x − 5)

 
     
Answer
     
     

Start with [y =frac{ 2}{ 3} x − 5] and multiply both sides by 3 to clear the fractions.

     

[3y = 2x − 15]

     

Finally, subtract 3y from both sides of the equation, then add 15 to both sides of the equation to obtain [15 = 2x − 3y], or equivalently, [2x − 3y = 15].

     
 
 
 
 

Exercise (PageIndex{26})

 

(y = frac{5}{6}x + 1)

 
 
 

Exercise (PageIndex{27})

 

(y = −frac{4}{ 5} x + 3)

 
     
Answer
     
     

Start with [y = −frac{4}{5} x + 3] and multiply both sides by 5 to clear the fractions. [5y = −4x + 15]

     

Finally, add 4x to both sides of the equation.

     

[4x + 5y = 15]

     
 
 
 
 

Ejercicio ( PageIndex {28} )

 

(y = −frac{3}{7} x + 2)

 
 
 

Ejercicio ( PageIndex {29} )

 

(y = −frac{2}{5} x − 3)

 
     
Answer
     
     

Start with [y = −frac{2}{5} x − 3] and multiply both sides by 5 to clear the fractions.

     

[5y = −2x − 15]

     

Finally, add 2x to both sides of the equation.

     

[2x + 5y = −15]

     
 
 
 
 

Ejercicio ( PageIndex {30} )

 

(y = −frac{1}{4}x + 2)

 
 
 

Exercise (PageIndex{31})

 

What is the x-intercept of the line?

 

屏幕快照 2019-09-21 上午12.30.41.png

 
     
Answer
     
     

The x-intercept is the location where the line crosses the x-axis.

     

Screen Shot 2019-09-28 at 10.54.22 PM.png

     

Therefore, the x-intercept is (−4, 0).

     
 
 
 
 

Exercise (PageIndex{32})

 

What is the y-intercept of the line?

 

屏幕快照 2019-09-21 上午12.31.40.png

 
 
 

Exercise (PageIndex{33})

 

What is the y-intercept of the line?

 

屏幕快照 2019-09-21 上午12.33.17.png

 
     
Answer
     
     

The y-intercept is the location where the line crosses the y-axis.

     

Screen Shot 2019-09-28 at 11.02.41 PM.png

     

Therefore, the y-intercept is (0, 4).

     
 
 
 
 

Exercise (PageIndex{34})

 

What is the x-intercept of the line?

 

屏幕快照 2019-09-21 上午12.35.02.png

 
 

In Exercises (PageIndex{35})-(PageIndex{40}), find the x- and y-intercepts of the linear function that is given in standard form. Use the intercepts to plot the graph of the line on a sheet of graph paper.

 
 

Exercise (PageIndex{35})

 

3x − 2y = 6

 
     
Answer
     
     

Set x = 0 in 3x − 2y = 6 to get −2y = 6 or y = −3. The y-intercept is (0, −3). Set y = 0 in 3x − 2y = 6 to get 3x = 6 or x = 2. The x-intercept is (2, 0). Plot the intercepts. The line through the intercepts is the required line.

     

Screen Shot 2019-09-28 at 11.03.33 PM.png

     
 
 
 
 

Exercise (PageIndex{36})

 

4x + 5y = 20

 
 
 

Exercise (PageIndex{37})

 

x − 2y = −2

 
     
Answer
     
     

Set x = 0 in x−2y = −2 to get −2y = −2 or y = 1. The y-intercept is (0, 1). Set y = 0 in x − 2y = −2 to get x = −2. The x-intercept is (−2, 0). Plot the intercepts. The line through the intercepts is the required line.

     

Screen Shot 2019-09-28 at 11.04.12 PM.png

     
 
 
 
 

Exercise (PageIndex{38})

 

6x + 5y = 30

 
 
 

Exercise (PageIndex{39})

 

2x − y = 4

 
     
Answer
     
     

Set x = 0 in 2x − y = 4 to get −y = 4 or y = −4. The y-intercept is (0, −4). Set y = 0 in 2x − y = 4 to get 2x = 4. The x-intercept is (2, 0). Plot the intercepts. The line through the intercepts is the required line.

     

Screen Shot 2019-09-28 at 11.04.57 PM.png

     
 
 
 
 

Exercise (PageIndex{40})

 

8x − 3y = 24

 
 
 

Exercise (PageIndex{41})

 

Sketch the graph of the horizontal line that passes through the point (3, −3). Label the line with its equation.

 
     
Answer
     
     

Every horizontal line has an equation of the form y = d. Since this line must pass through the point (3, −3), it follows that the equation is y = −3.

     

Screen Shot 2019-09-28 at 11.05.37 PM.png

     
 
 
 
 

Exercise (PageIndex{42})

 

Sketch the graph of the horizontal line that passes through the point (−9, 9). Label the line with its equation.

 
 
 

Exercise (PageIndex{43})

 

Sketch the graph of the vertical line that passes through the point (2, −1). Label the line with its equation.

 
     
Answer
     
     

Every vertical line has an equation of the form x = c. Since this line must pass through the point (2, −1), it follows that the equation is x = 2.

     

Screen Shot 2019-09-28 at 11.06.17 PM.png

     
 
 
 
 

Exercise (PageIndex{44})

 

Sketch the graph of the vertical line that passes through the point (15, −16). Label the line with its equation.

 
 

In Exercises (PageIndex{45})-(PageIndex{48}), find the domain and range of the given linear function.

 
 

Exercise (PageIndex{45})

 

f(x) = −37x − 86

 
     
Answer
     
     

The domain of every linear function is ((−infty,infty)). Since the slope of the graph of f is (−37 neq 0), the range is also ((−infty,infty)).

     
 
 
 
 

Exercise (PageIndex{46})

 

f(x) = 98

 
 
 

Exercise (PageIndex{47})

 

f(x) = −12

 
     
Answer
     
     

The domain of every linear function is ((−infty,infty)). Since f(x) = −12 for every x, the range is {−12}.

     
 
 
 
 

Exercise (PageIndex{48})

 

f(x) = −2x + 8

 
 

3.4 Exercises 

 

In Exercises (PageIndex{1})-(PageIndex{4}), perform each of the following tasks.

 

i. Draw the line on a sheet of graph paper with the given slope m that passes through the given point ((x_{0}, y_{0})).

 

ii. Estimate the y-intercept of the line.

 

iii. Usa la forma punto-pendiente para determinar la ecuación de la recta. Place your answer in slope-intercept form by solving for y. Compare the exact value of the y-intercept with the approximation found in part (ii).

 
 

Exercise (PageIndex{1})

 

m = 2/3 and ((x_{0}, y_{0}) = (−1, −1))

 
     
Answer
     
     

Plot the point P(−1, −1). To draw a line through P with slope m = 2/3, start at the point P, then move up 2 units and right 3 units to the point Q(2, 1). The line through the points P and Q is the required line.

     

Screen Shot 2019-09-28 at 11.32.55 PM.png

     

From the graph above, we would estimate the y-intercept as (0, −0.3). To find the equation of the line, substitute m = 2/3 and ((x_{0}, y_{0}) = (−1, −1)) into the point-slope form of the line.

     

[y − y0 = m(x − x0) \ y − (−1) = frac{2}{ 3} (x − (−1))\  y + 1 = frac{2}{ 3} (x + 1)]

     

To place this in slope-intercept form y = mx + b, solve for y.

     

[y = frac{2}{ 3} x + frac{2}{ 3} − 1\ y = frac{2}{ 3} x + frac{2}{ 3} − frac{3}{ 3} \ y = frac{2}{ 3} x − frac{1}{ 3}]

     

Comparing this result with y = mx + b, we see that the exact y-coordinate of the yintercept is b = −1/3, which is in close agreement with the approximation −0.3 found above.

     
 
 
 
 

Exercise (PageIndex{2})

 

m = -2/3 and ((x_{0}, y_{0}) = (1, −1))

 
 
 

Exercise (PageIndex{3})

 

m = -3/4 and ((x_{0}, y_{0}) = (−2, 3))

 
     
Answer
     
     

Plot the point P(−2, 3). To draw a line through P with slope m = −3/4, start at the point P, then move down 3 units and right 4 units to the point Q(2, 0). The line through the points P and Q is the required line.

     

Screen Shot 2019-09-28 at 11.36.18 PM.png

     

From the graph above, we would estimate the y-intercept as (0, 1.5). To find the equation of the line, substitute m = −3/4 and ((x_{0}, y_{0}) = (−2, 3)) into the point-slope form of the line.

     

[y − y_{0} = m(x − x_{0}) \ y − 3 = −frac{3}{ 4} (x − (−2)) \ y − 3 = −frac{3}{ 4} (x + 2)]

     

To place this in slope-intercept form y = mx + b, solve for y.

     

[y = −frac{3}{ 4} x − frac{3}{ 2} + 3 \ y = −frac{3}{ 4} x − frac{3}{ 2} + frac{6}{ 2} \ y = −frac{3}{ 4} x + frac{3}{ 2} ]

     

Comparing this result with y = mx + b, we see that the exact y-coordinate of the y-intercept is b = 3/2, which is in close agreement with the approximation 1.5 found above.

     
 
 
 
 

Exercise (PageIndex{4})

 

m = 2/5 and ((x_{0}, y_{0}) = (−3, -2))

 
 
 

Exercise (PageIndex{5})

 

Find the equation of the line in slope-intercept form that passes through the point (1, 3) and has a slope of 1.

 
     
Answer
     
     

Substitute 1 for m, 1 for (x_{1}), and 3 for (y_{1}) into the point-slope form (y − y_{1} = m(x − x_{1})) to obtain y − 3 = 1(x − 1). To place this in slope-intercept form y = mx + b, solve for y. [y = x − 1 + 3 \ y = x + 2]

     
 
 
 
 

Exercise (PageIndex{6})

 

Find the equation of the line in slope-intercept form that passes through the point (0, 2) and has a slope of 1/4.

 
 
 

Exercise (PageIndex{7})

 

Find the equation of the line in slope-intercept form that passes through the point (1, 9) and has a slope of −2/3.

 
     
Answer
     
     

Substitute −2/3 for m, 1 for x1, and 9 for y1 into the point-slope form (y − y_{1} = m(x − x_{1}))

     

to obtain [y − 9 = −frac{2}{3} (x − 1)].

     

To place this in slope-intercept form y = mx + b, solve for y.

     

[y = −frac{2}{3} x + frac{2}{3} + 9 \ y = −frac{2}{3} x + frac{2}{3} + frac{27}{3} \ y = −frac{2}{3} x + frac{29}{3}]

     
 
 
 
 

Exercise (PageIndex{8})

 

Find the equation of the line in slope-intercept form that passes through the point (1, 9) and has a slope of −3/4.

 
 

In Exercises (PageIndex{9})-(PageIndex{12}), perform each of the following tasks.

 

i. Set up a coordinate system on a sheet of graph paper and draw the line through the two given points.

 

ii. Usa la forma punto-pendiente para determinar la ecuación de la recta.

 

iii. Place the equation of the line in standard form Ax+By = C, where A, B, and C are integers and A > 0. Label the line in your plot with this result.

 
 

Exercise (PageIndex{9})

 

(−2, −1) and (3, 2)

 
     
Answer
     
     

Plot the points P(−2, −1) and Q(3, 2) and draw a line through them.

     

Screen Shot 2019-09-28 at 11.42.25 PM.png

     

Calculate the slope of the line through the points P and Q.

     

[m = dfrac{delta y }{delta x} = dfrac{2 − (−1)}{ 3 − (−2)} = dfrac{3}{5}]

     

Substitute m = 3/5 and ((x_{0}, y_{0}) = (−2, −1)) into the point-slope form of the line

     

[y − y_{0} = m(x − x_{0}) \  y − (−1) = dfrac{3}{5} (x − (−2))  \ y + 1 =dfrac{3}{5} (x + 2)]

     

Place this result in Standard form. First clear the fractions by multiplying by 5.

     

[y + 1 = dfrac{3}{5} x + dfrac{6}{5} \ 5y + 5 = 3x + 6 \ 3x − 5y = −1]

     

Hence, the standard form of the line is 3x − 5y = −1.

     
 
 
 
 

Exercise (PageIndex{10})

 

(−1, 4) and (2, −3)

 
 
 

Exercise (PageIndex{11})

 

(−2, 3) and (4, −3)

 
     
Answer
     
     

Plot the points P(−2, 3) and Q(4, −3) and draw a line through them.

     

Screen Shot 2019-09-29 at 10.46.51 AM.png

     

Calculate the slope of the line through the points P and Q.

     

[m = dfrac{delta y}{delta x} = dfrac{−3 − 3}{4 − (−2)}= dfrac{−6}{ 6} = −1]

     

Substitute m = −1 and ((x_{0}, y_{0}) = (−2, 3)) into the point-slope form of the line.

     

[y − y_{0} = m(x − x_{0}) \ y − 3 = −1(x − (−2)) \  y − 3 = −1(x + 2) ]

     

Place this result in Standard form.

     

[y − 3 = −x − 2 \ x + y = 1 ]

     

Hence, the standard form of the line is x + y = 1.

     
 
 
 
 

Exercise (PageIndex{12})

 

(−4, 4) and (2, −4)

 
 
 

Exercise (PageIndex{13})

 

Find the equation of the line in slope-intercept form that passes through the points (−5, 5) and (6, 8).

 
     
Answer
     
     

Substitute 5 for (y_{1}), 8 for (y_{2}), −5 for (x_{1}), and 6 for (x_{2}) into the slope formula to find the slope m:

     

[m = dfrac{y_{1} − y_{2}}{x_{1} − x_{2} }= dfrac{5 − 8}{−5 − 6} = dfrac{3}{11}]

     

Now substitute (dfrac{3}{11}) for m, −5 for x1, and 5 for y1 into the point-slope form

     

[y − y_{1} = m(x − x_{1})]

     

and then solve for y to obtain the equation

     

[y = dfrac{3}{ 11} x + dfrac{70}{ 11}]

     
 
 
 
 

Exercise (PageIndex{14})

 

Find the equation of the line in slope-intercept form that passes through the points (6, −6) and (9, −7).

 
 
 

Exercise (PageIndex{15})

 

Find the equation of the line in slope-intercept form that passes through the points (−4, 6) and (2, −4).

 
     
Answer
     
     

Substitute 6 for (y_{1}), −4 for (y_{2}), −4 for (x_{1}), and 2 for (x_{2}) into the slope formula to find the slope m:

     

[m = dfrac{ y_{1} − y_{2}}{  x_{1} − x_{2} }= dfrac{6 − (−4) }{−4 − 2} = dfrac{−5}{ 3}]

     

Now substitute (dfrac{−5}{ 3}) for m, −4 for x1, and 6 for (y_{1}) into the point-slope form

     

[y − y_{1} = m(x − x_{1})]

     

 and then solve for y to obtain the equation

     

[y = dfrac{−5}{3} x − dfrac{2}{ 3}]

     
 
 
 
 

Exercise (PageIndex{16})

 

Find the equation of the line in slope-intercept form that passes through the points (−1, 5) and (4, 4).

 
 

In Exercises (PageIndex{17})-(PageIndex{20}), perform each of the following tasks.

 

i. Draw the graph of the given linear equation on graph paper and label it with its equation.

 

ii. Determine the slope of the given equation, then use this slope to draw a second line through the given point P that is parallel to the first line.

 

iii. Estimate the y-intercept of the second line from your graph.

 

iv. Use the point-slope form to determine the equation of the second line. Place this result in slope-intercept form y = mx + b, then state the exact value of the y-intercept. Label the second line with the slope-intercept form of its equation.

 
 

Exercise (PageIndex{17})

 

2x + 3y = 6, P = (−2, −3)

 
     
Answer
     
     

Plot the points Q(0, 2) and R(3, 0) and draw a line through them as shown in (a) below. You can calculate the slope of this line from the graph, or you can use the slope formula as follows.

     

[m = dfrac{delta y}{delta x} = dfrac{0 − 2 }{3 − 0} = −dfrac{2}{ 3}]

     

The second line must be parallel to the first, so it must have the same slope; namely, m = −2/3. The second line must pass through the point P(−2, −3), so plot the point P. To get the right slope, start at the point P, then move 3 units to the right and 2 units down, as shown in (b). It would appear that this line crosses the y-axis near (0, −4.3).

     

Screen Shot 2019-09-29 at 11.13.00 AM.png

     

To find the equation of the second line, use the point slope form of the line and m = −2/3 and ((x_{0}, y_{0}) = (−2, −3)), as follows.

     

[y − y_{0} = m(x − x_{0}) \ y − (−3) = −dfrac{2}{ 3} (x − (−2))  \  y + 3 = −dfrac{2}{ 3}(x + 2)]

     

To place this in slope-intercept form y = mx + b, we must solve for y.

     

[y + 3 = −dfrac{2}{ 3}x − dfrac{4}{ 3} \  y = −dfrac{2}{ 3} x − dfrac{4}{ 3} − 3 \ y = −dfrac{2}{ 3} x − dfrac{4}{ 3} −dfrac{9}{ 3}\  y = −dfrac{2}{ 3} x − dfrac{13}{ 3}]

     

Hence, the equation in slope-intercept form is y = (−2/3)x − 13/3, making the exact y-coordinate of the y-intercept b = −13/3, which is in pretty close agreement (check on your calculator) with our estimate of −4.3.

     
 
 
 
 

Exercise (PageIndex{18})

 

3x − 4y = 12, P = (−3, 4)

 
 
 

Exercise (PageIndex{19})

 

x + 2y = −4, P = (3, 3)

 
     
Answer
     
     

Plot the points Q(−4, 0) and R(0, −2) and draw a line through them as shown in (a) below. You can calculate the slope of this line from the graph, or you can use the slope formula as follows.

     

[m = dfrac{delta y}{delta x} = dfrac{−2 − 0 }{0 − (−4)} = −dfrac{1}{2}]

     

The second line must be parallel to the first, so it must have the same slope; namely, m = −1/2. The second line must pass through the point P(3, 3), so plot the point P. To get the right slope, start at the point P, then move 1 unit downward and 2 units to the right, as shown in (b). It would appear that this line crosses the y-axis near (0, 4.5).

     

Screen Shot 2019-09-29 at 11.19.41 AM.png

     

To find the equation of the second line, use the point slope form of the line and m = −1/2 and ((x_{0}, y_{0}) = (3, 3)), as follows.

     

[y − y_{0} = m(x − x_{0}) \  y − 3 = −dfrac{1}{ 2} (x − 3) ]

     

To place this in slope-intercept form y = mx + b, we must solve for y.

     

[− 3 = −dfrac{1}{ 2} x + dfrac{3}{ 2} \ y = −dfrac{1}{ 2} x + dfrac{3}{ 2} + 3 \ y = −dfrac{1}{ 2} x + dfrac{3}{ 2} + dfrac{6}{ 2} \ y = −dfrac{1}{ 2} x + dfrac{9}{ 2}]

     

Hence, the equation in slope-intercept form is y = (−1/2)x + 9/2, making the exact y-coordinate of the y-intercept b = 9/2, which is in pretty close agreement (check on your calculator) with our estimate of 4.5.

     
 
 
 
 

Exercise (PageIndex{20})

 

5x + 2y = 10, P = (−3, −5)

 
 

In Exercises (PageIndex{21})-(PageIndex{24}), perform each of the following tasks.

 

i. Draw the graph of the given linear equation on graph paper and label it with its equation.

 

ii. Determine the slope of the given equation, then use this slope to draw a second line through the given point P that is prependicular to the first line.

 

iii. Use the point-slope form to determine the equation of the second line. Place this result in standard form Ax+By = C, where A, B, C are integers and A > 0. Label the second line with this standard form of its equation.

 
 

Exercise (PageIndex{21})

 

x − 2y = −2, P = (3, −4)

 
     
Answer
     
     

Let x = 0 in x − 2y = −2. Then −2y = −2 and y = 1. This calculation gives us the y-intercept R(0, 1). Let y = 0 in x − 2y = −2 and x = −2. This gives us the x-intercept Q(−2, 0). Plot the points Q(−2, 0) and R(0, 1) and draw a line through them as in (a) below.

     

Screen Shot 2019-09-29 at 11.24.47 AM.png

     

You can calculate the slope of the first line from the graph or you can obtain it with the slope formula as follows.

     

[m1 = dfrac{delta y}{delta x} = dfrac{1 − 0}{ 0 − (−2)} = dfrac{1}{ 2}]

     

The second line is perpendicular to this first line, so the slope of the second line must be the negative reciprocal of the slope of the first line; i.e., the slope of the second line should be (m = −1/m_{2} = −1/(1/2)), or m = −2. To draw a line through the point P(3, −4) that is perpendicular to the line in (a), first plot the point P(3, −4), then move upward 2 units and to the left 1 unit as shown in (b). This gives us a line through P(3, −4) with slope m = −2, so this line will be perpendicular to the first line. To find the equation of the perpendicular line, substitute m = −2 and ((x_{0}, y_{0}) = (3, −4)) into the point-slope formula, then place the resulting equation in standard form.

     

[y − y_{0} = m(x − x_{0}) \ y − (−4) = −2(x − 3) \ y + 4 = −2x + 6 \ 2x + y = 2 ]

     

Thus, the equation of the line that passes through P(3, −4) and is perpendicular to the line x − 2y = −2 is 2x + y = 2.

     
 
 
 
 

Exercise (PageIndex{22})

 

3x + y = 3, P = (−3, −4)

 
 
 

Exercise (PageIndex{23})

 

x − 2y = 4, P = (−3, 3)

 
     
Answer
     
     

Set x = 0 in x − 2y = 4 to obtain −2y = 4. Hence, y = −2 and the y-intercept is Q(0, −2). Set y = 0 in x − 2y = 4 to obtain x = 4. Hence, the x-intercept is R(4, 0). Plot Q(0, −2) and R(4, 0) and draw a line through them as in (a) below.

     

Screen Shot 2019-09-29 at 11.28.56 AM.png

     

You can calculate the slope of the first line from the graph or you can obtain it with the slope formula as follows.

     

[m_{1} = dfrac{delta y}{delta x} = dfrac{0 − (−2)}{ 4 − 0 }= dfrac{1}{2}]

     

The second line is perpendicular to this first line, so the slope of the second line must be the negative reciprocal of the slope of the first line; i.e., the slope of the second line should be m = −1/m2 = −1/(1/2), or m = −2. To draw a line through the point P(−3, 3) that is perpendicular to the line in (a), first plot the point P(−3, 3), then move downward 2 units and to the right 1 unit as shown in (b). This gives us a line through P(−3, 3) with slope m = −2, so this line will be perpendicular to the first line. To find the equation of the perpendicular line, substitute m = −2 and ((x_{0}, y_{0}) = (−3, 3)) into the point-slope formula, then place the resulting equation in standard form.

     

[y − y_{0} = m(x − x_{0}) \ y − 3 = −2(x − (−3)) \ y − 3 = −2(x + 3) \ y − 3 = −2x − 6 \ 2x + y = −3]

     

Thus, the equation of the line that passes through P(−3, 3) and is perpendicular to the line x − 2y = 4 is 2x + y = −3.

     
 
 
 
 

Exercise (PageIndex{24})

 

x − 4y = 4, P = (−3, 4)

 
 
 

Exercise (PageIndex{25})

 

Find the equation of the line in slope-intercept form that passes through the point (7, 8) and is parallel to the line x − 5y = 4.

 
     
Answer
     
     

First solve x − 5y = 4 for y to get (y = dfrac{1 }{5} x − dfrac{4}{5} ) The slope of this line is (dfrac{1}{5}). Therefore, every parallel line also has slope (dfrac{1}{5}).

     

Now to find the equation of the parallel line that passes through the point (7, 8), substitute (dfrac{1}{5}) for m, 7 for (x_{1}), and 8 for (y_{1}) into the point-slope form

     

[y − y_{1} = m(x − x_{1})] to obtain

     

[y − 8 = dfrac{1}{5}(x − 7)]. Then solve for y:

     

[y = dfrac{1}{5} x − dfrac{7}{5} + 8 \ y = dfrac{1}{5}x − dfrac{7}{5} +dfrac{40}{5} y = dfrac{1}{5} x + dfrac{33}{5}] 

     
 
 
 
 

Exercise (PageIndex{26})

 

Find the equation of the line in slope-intercept form that passes through the point (3, −7) and is perpendicular to the line 7x − 2y = −8.

 
 
 

Exercise (PageIndex{27})

 

Find the equation of the line in slope-intercept form that passes through the point (1, −2) and is perpendicular to the line −7x + 5y = 4.

 
     
Answer
     
     

First solve −7x + 5y = 4 for y to get

     

[y = dfrac{7}{5} x + 4 5]

     

The slope of this line is dfrac{7}{5}. Therefore, every perpendicular line has slope −(dfrac{7}{5}) (the negative reciprocal of (dfrac{7}{5}). Now to find the equation of the perpendicular line that passes through the point (1, −2), substitute −dfrac{5}{7} for m, 1 for (x_{1}), and −2 for (y_{1}) into the point-slope form

     

[y − y1 = m(x − x1)] to obtain [y − (−2) = −dfrac{5}{7} (x − 1)]. Then solve for y:

     

[y = −dfrac{5}{7}x +dfrac{5}{7} − 2 \ y = −dfrac{5}{7} x + dfrac{5}{7} − dfrac{14}{7} \ y = −dfrac{5}{7} x − dfrac{9}{7}]

     
 
 
 
 

Ejercicio ( PageIndex {28} )

 

Find the equation of the line in slope-intercept form that passes through the point (4, −9) and is parallel to the line 9x + 3y = 5.

 
 
 

Ejercicio ( PageIndex {29} )

 

Find the equation of the line in slope-intercept form that passes through the point (2, −9) and is perpendicular to the line −8x + 3y = 1.

 
     
Answer
     
     

First solve −8x + 3y = 1 for y to get [y = dfrac{8}{3} x + dfrac{1}{3} ]

     

The slope of this line is (dfrac{8}{3}). Therefore, every perpendicular line has slope (−dfrac{3}{8}) (the negative reciprocal of (dfrac{8}{3})). Now to find the equation of the perpendicular line that passes through the point (2, −9), substitute (−dfrac{3}{8}) for m, 2 for (x_{1}), and −9 for (y_{1}) into the point-slope form [y − y1 = m(x − x1)]

     

to obtain [y − (−9) = −dfrac{3}{8} (x − 2)]

     

Then solve for y:

     

[y = −dfrac{3}{8} x + dfrac{3}{4} − 9 \ y = −dfrac{3}{8} x + dfrac{3}{4}− dfrac{36}{4} \ y = −dfrac{3}{8} x − dfrac{33}{4}]

     
 
 
 
 

Ejercicio ( PageIndex {30} )

 

Find the equation of the line in slope-intercept form that passes through the point (−7, −7) and is parallel to the line 8x + y = 2.

 
     
Answer
     
     

Add texts here. Do not delete this text first.

     
 
 
 
 

Exercise (PageIndex{31})

 

A ball is thrown vertically upward on a distant planet. After 1 second, its velocity is 100 meters per second. After 5 seconds, the velocity is 50 meters per second. Assume that the velocity v of the ball is a linear function of the time t.

 

a) On graph paper, sketch the graph of the velocity v versus the time t. Assume that the velocity is the dependent variable and place it on the vertical axis.

 

b) Determine the slope of the line, including its units, then give a real world explanation of the meaning of this slope.

 

c) Determine an equation that models the velocity v of the ball as a function of time t.

 

d) Determine the time it takes the ball to reach its maximum height.

 
     
Answer
     
     

a) At 1 second, the speed is 100 meters per second. This is the point (1, 100) in the plot below. At 5 seconds, the speed is 50 meters per second. This is the point (5, 50) in the plot below.

     

Screen Shot 2019-10-02 at 3.15.51 PM.png

     

b) We’ll keep the units in our slope calculation to provide real-world meaning for the rate.

     

[m = dfrac{delta v}{delta t} = dfrac{50 m/s − 100 m/s}{ 5 s − 1 s} = −12.5 (m/s)/s.]

     

That is, the slope is (−12.5 m/s^2). This is acceleration, the rate at which the speed is changing with respect to time. Every second, the speed decreases by 12.5 meters per second.

     

c) We use the point-slope form of the line, namely

     

[y − y_{0} = m(x − x_{0})]

     

However, v and t are taking the place of y and x, respectively, so the equation becomes [v − v_{0} = m(t − t_{0})], Now, substitute the slope m = −12.5 and the point ((t_{0}, v_{0}) = (1, 100)) to obtain [v − 100 = −12.5(t − 1)]. Solve this for v, obtaining [v − 100 = −12.5t + 12.5 \ v = −12.5t + 112.5]. Equivalently, we could use function notation and write [v(t) = −12.5t + 112.5].

     

d) When the ball reaches its maximum height, its speed will equal zero. Consequently, to find the time of this event, we must solve v(t) = 0. Replace v(t) with −12.5t + 112.5 and solve for t. [−12.5t + 112.5 = 0 \ −12.5t = −112.5 \ t = 9]. Hence, it takes 9 seconds for the ball to reach its maximum height.

     
 
 
 
 

Exercise (PageIndex{32})

 

A ball is thrown vertically upward on a distant planet. After 2 seconds, its velocity is 320 feet per second. After 8 seconds, the velocity is 200 feet per second. Assume that the velocity v of the ball is a linear function of the time t.

 

a) On graph paper, sketch the graph of the velocity v versus the time t. Assume that the velocity is the dependent variable and place it on the vertical axis.

 

b) Determine the slope of the line, including its units, then give a real world explanation of the meaning of this slope.

 

c) Determine an equation that models the velocity v of the ball as a function of time t.

 

d) Determine the time it takes the ball to reach its maximum height.

 
 
 

Exercise (PageIndex{33})

 

An automobile is traveling down the autobahn and the driver applies its brakes. After 2 seconds, the car’s speed is 60 km/h. After 4 seconds, the car’s speed is 50 km/h.

 

a) On graph paper, sketch the graph of the velocity v versus the time t. Assume that the velocity is the dependent variable and place it on the vertical axis.

 

b) Determine the slope of the line, including its units, then give a real world explanation of the meaning of this slope.

 

c) Determine an equation that models the velocity v of the automobile as a function of time t. d) Determine the time it takes the automobile to stop.

 
     
Answer
     
     

a) After 2 seconds, the speed of the car is 60 km/h. This is the point (2, 60). After 4 seconds, the car’s speed is 50 km/h. This is the point (4, 50). Plot these two points and draw a line through them, as shown in the plot below.

     

Screen Shot 2019-10-02 at 3.19.44 PM.png

     

b) We’ll keep the units in our slope calculation to provide real-world meaning for the rate.

     

[m = dfrac{delta v}{delta t} = dfrac{50 km/h − 60 km/h}{ 4 s − 2 s }= −5 (km/h)/s]

     

That is, the slope is −5 (km/h)/s. This is acceleration, the rate at which the speed is changing with respect to time. Every second, the speed decreases by 5 kilometers per hour.

     

c) We use the point-slope form of the line, namely [y − y_{0} = m(x − x_{0})], However, v and t are taking the place of y and x, respectively, so the equation becomes [v − v_{0} = m(t − t_{0})], Now, substitute the slope m = −5 and the point ((t_{0}, v_{0}) = (2, 60)) to obtain [v − 60 = −5(t − 2)]. Solve this for v, obtaining [v − 60 = −5t + 10 \  v = −5t + 70]. Equivalently, we could use function notation and write [v(t) = −5t + 70].

     

d) To find the time it takes the car to stop, we must determine the time t so that [v(t) = 0]. Replace v(t) with −5t + 70 and solve for t.

     

[−5t + 70 = 0\  −5t = −70 \ t = 14]. Hence, it takes 14 seconds for the to brake to a stop.

     
 
 
 
 

Exercise (PageIndex{34})

 

An automobile is traveling down the autobahn and its driver steps on the accelerator. After 2 seconds, the car’s velocity is 30 km/h. After 4 seconds, the car’s velocity is 40 km/h.

 

a) On graph paper, sketch the graph of the velocity v versus the time t. Assume that the velocity is the dependent variable and place it on the vertical axis.

 

b) Determine the slope of the line, including its units, then give a real world explanation of the meaning of this slope.

 

c) Determine an equation that models the velocity v of the automobile as a function of time t.

 

d) Determine the speed of the vehicle after 8 seconds.

 
 
 

Exercise (PageIndex{35})

 

Suppose that the demand d for a particular brand of teakettle is a linear function of its unit price p. When the unit price is fixed at $30, the demand for teakettles is 100. This means the public buys 100 teakettles. If the unit price is fixed at $50, then the demand for teakettles is 60.

 

a) On graph paper, sketch the graph of the demand d versus the unit price p. Assume that the demand is the dependent variable and place it on the vertical axis.

 

b) Determine the slope of the line, including its units, then give a real world explanation of the meaning of this slope.

 

c) Determine an equation that models the demand d for teakettles as a function of unit price p.

 

d) Compute the demand if the unit price is set at $40.

 
     
Answer
     
     

a) When the unit price is $30, the demand is 100 teakettles. This is the point (30, 100). When the unit price is $50, the demand is 60 teakettles. This is the point (50, 60). Plot the points (30, 100) and (50, 60) and draw a line through them, as shown in the plot below.

     

Screen Shot 2019-10-02 at 3.23.12 PM.png

     

b) We’ll keep the units in our slope calculation to provide real-world meaning for the rate.

     

[m = dfrac{delta d}{delta p} = dfrac{60 teakettles − 100 teakettles}{ 50 dollars − 30 dollars } = −2 teakettles/dollar]

     

That is, the slope is −2 teakettles/dollar. This is the rate at which the demand is changing with respect to the unit price. For every increase in the unit price of one dollar, the demand is lowered by 2 teakettles (2 fewer teakettles are bought).

     

c) We use the point-slope form of the line, namely[ y − y_{0} = m(x − x_{0})], However, d and p are taking the place of y and x, respectively, so the equation becomes

     

[d − d_{0} = m(p − p_{0})]

     

Now, substitute the slope m = −2 and the point ((p_{0}, d_{0}) = (30, 100)) to obtain [d − 100 = −2(p − 30)]. Solve this for d, obtaining

     

[d − 100 = −2p + 60 \ d = −2p + 160]

     

Equivalently, we could use function notation and write d(p) = −2p + 160.

     

d) To determine the demand if the unit price is $40, compute [ d(40) = −2(40) + 160 = 80]

     

Hence, the demand is for 80 teakettles if the price per kettle is set at $40.

     
 
 
 
 

Exercise (PageIndex{36})

 

It’s perfect kite-flying weather on the coast of Oregon. Annie grabs her kite, climbs up on the roof of her two story home, and begins playing out kite string. In 10 seconds, Annie’s kite is 120 feet above the ground. After 20 seconds, it is 220 feet above the ground. Assume that the height h of the kite above the ground is a linear function of the amount of time t that has passed since Annie began playing out kite string.

 

a) On graph paper, sketch the graph of the height h of the kite above ground versus the time t . Assume that the height is the dependent variable and place it on the vertical axis.

 

b) Determine the slope of the line, including its units, then give a real world explanation of the meaning of this slope.

 

c) Determine an equation that models the height h of the kite as a function of time t.

 

d) Determine the height of the kite after 20 seconds.

 

e) Determine the height of Annie’s second story roof above ground.

 
 

3.5 Exercises 

 
 

Exercise (PageIndex{1})

 

The following set of data about revolving consumer credit (debt) in the United States is from Google.com. This is primarily made up of credit card debt, but also includes other consumer non-mortgage credit, like those offered by commercial banks, credit unions, Sallie Mae, and the federal government.

                                                                                                                                                                                                                                                                                                                                                                                                               
Year yrs x after 2001 all revolving credit C in billions of $
2001 0 721.0
2002 1 741.2
2003 2 759.3
2004 3 786.1
2005 4 805.4
 

a) Set up a coordinate system on graph paper, placing the credit C on the vertical axis, and the years x after 2001 on the horizontal axis. Label and scale each axis appropriately. Draw what you feel is the line of best fit. Remember to draw all lines with a ruler.

 

b) Select two points on your line of best fit that are not from the data table above. Use these two points to determine the slope of the line. Include units with your answer. Write a sentence or two explaining the real world significance of the slope of the line of best fit.

 

c) Use one of the two points on the line and the slope to determine the equation of the line of best fit in point-slope form. Use C and x for the dependent and independent variables, respectively. Solve the resulting equation for C and write your result using function notation.

 

d) Use the equation developed in part (c) to predict the revolving credit debt in the year 2008.

 

e) If the linear trend predicted by the line of best fit continues, in what year will the revolving credit debt reach 1.0 trillion dollars?

 
     
Answer
     
     

a) Scale and label the axes, plot the points, then draw the line of best fit.

     

Screen Shot 2019-10-02 at 3.55.10 PM.png

     

b) Select two points, P(1.4, 750) and Q(2.8, 780) that are points on the line but not points in the original data table.

     

Screen Shot 2019-10-02 at 3.55.52 PM.png

     

Use these two points to calculate the slope of the line of best fit.

     

[m = dfrac{delta C}{delta x} = dfrac{780 − 750}{ 2.8 − 1.4}] billion dollars/year

     

Using a calculator, the slope is approximately 21.42 billion dollars per year. What this means is that the credit debt is increasing at a rate of 21.42 billion dollars per year. c) To find the equation of the line, use the point P(1.4, 750) and the slope m = 21.42 in the point-slope form of the line.

     

[y − y_{0} = m(x − x_{0}) \ y − 750 = 21.42(x − 1.4)]

     

Replace y and x with C and x, respectively, then solve for C in terms of x.

     

[C − 750 = 21.42(x − 1.4) \ C − 750 = 21.42x − 29.988 \ C = 21.42x + 720.012]

     

In function notation, C(x) = 21.42x + 720.012. Note: Answers will vary somewhat due to the subjective nature of drawing the line of best fit and picking points on the line.

     

d) The year 2008 gives x = 2008−2001 = 7 years since 2001. Hence, to find the credit debt in 2008, we use C(x) = 21.42x + 720.012 and evaluate

     

[C(7) = 21.42(7) + 720.012 = 869.952]

     

 Hence, the credit debt in 2008 will be approximately 869 billion dollars.

     

e) A trillion dollars is 1000 billion dollars. Hence, to find when the credit debt is 1000 billion dollars, we must solve C(x) = 1000 for x.

     

[C(x) = 1000 \ 21.42x + 720.012 = 1000 \  21.42x = 279.988 \ x approx 13.07] Hence, the credit debt will reach a trillion dollars approximately 13 years after the year 2001, or a little bit into the year 2014.

     
 
 
 
 

Exercise (PageIndex{2})

 

The following set of data about non-revolving credit (debt) in the United States is from Google.com. The largest component of non-revolving credit is automobile loans, but it is also includes student loans and other defined-term consumer loans.

                                                                                                                                                                                                                                                                                                                                                                                                   
Year yrs x after 2001 Nonrevolving debt D in billions of $
2001 0 1121.3
2002 1 1184.1
2003 2 1247.3
2004 3 1305.0
2005 4 1342.3
 

a) Set up a coordinate system on graph paper, placing the non-revolving credit debt D on the vertical axis, and the years x after 2001 on the horizontal axis. Label and scale each axis appropriately. Draw what you feel is the line of best fit. Remember to draw all lines with a ruler.

 

b) Select two points on your line of best fit that are not from the data table above. Use these two points to determine the slope of the line. Include units with your answer. Write a sentence or two explaining the real world significance of the slope of the line of best fit.

 

c) Use one of the two points on the line and the slope to determine the equation of the line of best fit in pointslope form. Use D and x for the dependent and independent variables, respectively. Solve the resulting equation for D and write your result using function notation.

 

d) Use the equation developed in part (c) to predict the non-revolving credit debt in the year 2008. e) If the linear trend predicted by the line of best fit continues, in what year will the non-revolving credit debt reach 2.0 trillion dollars?

 
 
 

Exercise (PageIndex{3})

 

 

 

According to the U.S. Bureau of Transportation (www.bts.gov), retail sales of new cars declined every year from 2000- 2004, as shown in the following table.

                                                                                                                                                                                                                                                                                                                                                                                                   
Year yrs x after 2000 Sales S in thousands
2001 0 8847
2002 1 8423
2003 2 8103
2004 3 7610
2005 4 7506
 

a) Set up a coordinate system on graph paper, placing the sales S on the vertical axis, and the years x after 2000 on the horizontal axis. Label and scale each axis appropriately. Draw what you feel is the line of best fit. Remember to draw all lines with a ruler.

 

b) Select two points on your line of best fit that are not from the data table above. Use these two points to determine the slope of the line. Include units with your answer. Write a sentence or two explaining the real world significance of the slope of the line of best fit.

 

c) Use one of the two points on the line and the slope to determine the equation of the line of best fit in point-slope form. Use S and x for the dependent and independent variables, respectively. Solve the resulting equation for S and write your result using function notation.

 

d) Use the equation developed in part (c) to predict sales in the year 2006.

 

e) If the linear trend predicted by the line of best fit continues, when will sales drop to 7 million cars per year?

 
     
Answer
     
     

a) Scale and label the axes, plot the points, then draw the line of best fit.

     

Screen Shot 2019-10-02 at 3.59.38 PM.png

     

b) Select two points, P(1.4, 8300) and Q(4.0, 7400) that are points on the line but not points in the original data table.

     

Screen Shot 2019-10-02 at 4.01.04 PM.png

     

Use these two points to calculate the slope of the line of best fit.

     

[m = dfrac{delta C }{delta x} = dfrac{8300 − 7400 }{4.0 − 1.4}] thousand cars/year Using a calculator, the slope is approximately −346.15 thousand cars per year. What this means is that sales of new cars is decreasing at an approximate rate of 346 thousand cars per year.

     

c) To find the equation of the line, use the point P(1.4, 8300) and the slope m = −346.15 in the point-slope form of the line.

     

[y − y_{0} = m(x − x_{0}) \ y − 8300 = −346.15(x − 1.4)]

     

Replace y and x with S and x, respectively, then solve for S in terms of x.

     

[S − 8300 = −346.15(x − 1.4) \ S − 8300 = −346.15x + 484.61 \ S = −346.15x + 8784.61]

     

In function notation, S(x) = −346.15x + 8784.61. Note: Answers will vary somewhat due to the subjective nature of drawing the line of best fit and picking points on the line.

     

d) To determine the sales in 2006, evaluate S(x) = −346.15x+8784.61 at x = 2006− 2000 = 6.

     

[S(6) = −346.15(6) + 8784.61 = 6707.71]

     

Thus, the sales will be approximately 6,707,710 new cars.

     

e) To determine when sales of new cars will drop to 7 million, we must solve

     

[S(x) = 7000. −346.15x + 8784.61 = 7000 \ −346.15x = −1784.61 \ x approx 5.16]

     

Hence, sales will reach 7 million cars somewhere in the year 2005-2006.

     
 
 
 
 

Exercise (PageIndex{4})

 

The following table shows total midyear population of the world according to the U.S. Census Bureau, (www.census.gov) for recent years.

                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                           
Year yrs x after 2000 Population P in billions
2000 0 6.08
2001 1 6.16
2002 2 6.23
2003 3 6.30
2004 4 6.38
2005 5 6.45
2006 6 6.53
 

 a) Set up a coordinate system on graph paper, placing the population P on the vertical axis, and the years x after 2000 on the horizontal axis. Label and scale each axis appropriately. Draw what you feel is the line of best fit. Remember to draw all lines with a ruler.

 

b) Select two points on your line of best fit that are not from the data table above. Use these two points to determine the slope of the line. Include units with your answer. Write a sentence or two explaining the real world significance of the slope of the line of best fit.

 

c) Use one of the two points on the line and the slope to determine the equation of the line of best fit in point-slope form. Use P and x for the dependent and independent variables, respectively. Solve the resulting equation for P and write your result using function notation.

 

d) Use the equation developed in part (c) to predict the population in 2010.

 

e) If the linear trend predicted by the line of best fit continues, when will world population reach 7 billion?

 
 
 

Exercise (PageIndex{5})

 

The following table shows an excerpt from the U.S. Census Bureau’s 2005 data (www.census.gov) on annual sales of new homes in the United States.

                                                                                                                                                                                                                           
Price Range (thousands of $) Number sold (thousands)
150 − 199 246
200 − 249 200
250 − 299 152
 

We cannot use price ranges as coordinate values (we must have single values), so we replace each price range in the table with a single price in the middle of the range–the average value of a home in that range. This gives us the following modified table:

                                                                                                                                                                                                                           
Avg Price P (thousands of $) Number sold N (thousands)
175 246
225 200
275 152
 

We can now plot the data on a coordinate system.

 

a) Enter the data into your calculator and make a scatter plot. Copy it down onto your paper, labeling appropriately.

 

b) Use your calculator to determine a line of best fit. This is called a linear demand function, because it allows you to predict the demand for houses with a certain price. Write it using function notation and round to the nearest thousandth. Graph it on your calculator and copy it onto your coordinate system.

 

c) Use the linear demand function to predict annual sales of homes priced at $200, 000. Try to use the TABLE feature on your calculator to make this prediction.

 
     
Answer
     
     

a) Select STAT, then 1:Edit and enter the data as shown in (a). Select 2nd STAT PLOT, then turn turn on Plot1 as shown in (b). Select scatterplot, choose the lists you used for the data, then a marker, as shown in (b). Finally, select 9:ZoomStat from the ZOOM menu to produce the plot in (c).

     

Screen Shot 2019-10-02 at 4.04.51 PM.png

     

b) Push the STAT button, right-arrow to the CALC menu, then select 4:LinReg(ax+b), followed by L1, a comma, L2, a comma, and Y1 from the VARS menu (select Y-VARS then 1:Function submenus). The result is shown in (a). Press ENTER to produce the calculation shown in (b). This is the line of best fit. This procedure also stores the line of best fit in Y1, so pushing the GRAPH button produces the line of best fit shown in (c).

     

Screen Shot 2019-10-02 at 4.05.52 PM.png

     

In (b) above, we see that the number sold N, as a function of average price P is given by the linear function N(P) = −0.94P + 410.833.

     

c) If all went well in part (b), then we can press 2nd TBL SET and set up the table parameters as shown in (g). Make sure the independent variable is set to ASK. Select 2nd TABLE, then enter 200 (which represents an average price p = $200, 000) as shown in (h).

     

Screen Shot 2019-10-02 at 4.07.07 PM.png

     

Thus, N(200) = 222.83, so approximately 222.83 thousand or 222, 830 homes are sold at the average price of $200,000.

     
 
 
 

 

 
 

Exercise (PageIndex{6})

 

The following table shows data from the National Association of Homebuilders (www.nahb.org), indicating the median price of new homes in the United States.

                                                                                                                                                                                                                                                                                                                                                                           
Year Median Price (thousands of $)
2000 169
2001 175
2002 188
2003 195
2004 221
2005 238
 

a) Enter the data into your calculator and make a scatter plot. Copy it down onto your paper, labeling appropriately.

 

b) Use your calculator to determine a line of best fit that can be used to predict the median price of new homes in future years. Write it using function notation. Graph it on your calculator and copy it onto your coordinate system.

 

c) Use the linear demand function to predict the median price of a new home in 2010. Try to use the TABLE feature on your calculator to make this prediction.

 

d) Looking at the graph, do you think the linear demand function models the actual data points well? If not, why not? What does this mean about the prediction you made in part (c)?

 
 
 

Exercise (PageIndex{7})

 

Jim is hanging blocks of various mass on a spring in the physics lab. He notices that the spring will stretch further if he adds more mass to the end of the spring. He is soon convinced that the distance the spring will stretch depends on the amount of mass attached to it. He decides to take some measurements. He records the amount of mass attached to the end of the spring and then measures the distance that the spring stretched. Here is Jim’s data.

                                                                                                                                                                                                                                                                                                                                                              
Mass (grams) Distance Stretched (cm)
50 1.2
100 1.9
150 3.1
200 4.0
250 4.8
300 6.2
 

a) Enter the data into your calculator and make a scatter plot. Copy it down onto your paper, labeling appropriately.

 

b) Use your calculator to determine a line of best fit that can be used to predict the distance the spring stretches. Write it using function notation. Graph it on your calculator and copy it onto your coordinate system.

 

c) Use the function from part (c) to predict the distance the spring will stretch if 175 grams is attached to the spring. Try to use the TABLE feature on your calculator to make this prediction.

 
     
Answer
     
     

a) Select STAT, then 1:Edit and enter the data as shown in (a). Select 2nd STAT PLOT, then turn turn on Plot1 as shown in (b). Select scatterplot, choose the lists you used for the data, then a marker, as shown in (b). Finally, select 9:ZoomStat from the ZOOM menu to produce the plot in (c).

     

Screen Shot 2019-10-02 at 4.08.09 PM.png

     

b) Push the STAT button, right-arrow to the CALC menu, then select 4:LinReg(ax+b), followed by L1, a comma, L2, a comma, and Y1 from the VARS menu (select Y-VARS then 1:Function submenus). The result is shown in (a). Press ENTER to produce the calculation shown in (b). This is the line of best fit. This procedure also stores the line of best fit in Y1, so pushing the GRAPH button produces the line of best fit shown in (c).

     

Screen Shot 2019-10-02 at 4.09.07 PM.png

     

In (b) above, we see that the distance stretched d, as a function of mass m is given by the linear function d(m) = 0.01977m + 0.07333.

     

c) If all went well in part (b), then we can press 2nd TBL SET and set up the table parameters as shown in (g). Make sure the independent variable is set to ASK. Select 2nd TABLE, then enter 175 (which represents a mass of 175 grams) as shown in (h).

     

Screen Shot 2019-10-02 at 4.10.44 PM.png

     

Thus, d(175) = 3.5333, so the spring stretches approximately 3.53 centimeters when a mass of 175 grams is attached.

     
 
 
 
 

Exercise (PageIndex{8})

 

Dave and Melody are lab partners in Tony Sartori’s afternoon chemistry lab. Professor Sartori has prepared an experiment to help them discover the relationship between the Celsius and Fahrenheit temperature scales. The experiment consists of a beaker full of ice and two thermometers, one calibrated in the Fahrenheit scale, the other in the Celsius scale. Dave and Melody use a Bunsen burner to heat the beaker, eventually bringing the water in the beaker to the boiling point. Every few minutees they make two temperature readings, one in Fahrenheit, one in Celsius. The data that they record during the laboratory session follows.

                                                                                                                                                                                                                                                                                                                                                                                                                           
Celsius Fahrenheit
4.0 39
18 65
30 85
51 122
70 159
85 186
100 210
 

a) Enter the data into your calculator and make a scatter plot. Copy it down onto your paper, labeling appropriately.

 

b) Use your calculator to determine a line of best fit that can be used to predict the Fahrenheit temperature as a function of the Celsius temperature. Write it using function notation. Graph it on your calculator and copy it onto your coordinate system.

 

c) Use the function from part (c) to predict the Fahrenheit temperature if the Celsius temperature is 40. Try to use the TABLE feature on your calculator to make this prediction.

 

d) Use the function from part (c) to predict the Celsius temperature if the Fahrenheit temperature is 100. 

 
 
 

Exercise (PageIndex{9})

 

The following table shows data on home sales at the Mendocino Coast in 2005.

                                                                                                                                                                                                                           
Price Range (thousands of $) Number sold (thousands)
200 − 299 14
300 − 399 55
400 − 499 62
 

We cannot use price ranges as coordinate values (we must have single values), so we replace each price range in the table with a single price in the middle of the range–the average value of a home in that range. This gives us the following modified table:

                                                                                                                                                                                                                           
Avg Price P (thousands of $) Number sold N (thousands)
250 14
350 55
450 62
 

We can now plot the data on a coordinate system.

 

a) Enter the data into your calculator and make a scatter plot. Copy it down onto your paper, labeling appropriately.

 

b) Use your calculator to determine a line of best fit. Write it using function notation and round to the nearest thousandth. Graph it on your calculator and copy it onto your coordinate system.

 

c) Use the linear function to predict the sales for houses in the price range $500, 000− $599, 000. Use the average price of $550, 000 for this estimate.

 

d) The actual number of houses sold in the price range $500, 000 − $599, 000 was 41. Plot this as a point on your coordinate system and compare it to your linear function model’s prediction. Notice that this actual value is pretty different from the prediction.

 

e) What this means is that a linear model is not very good for the data for home sales! Draw a simple curve that goes through each of the data points. Notice that it does not very closely resemble the shape of a line! More sophisticated functions are required to model this example–such as quadratic functions, which we study in a later chapter. The moral of the story here is that not every data set can be modeled linearly!

 
     
Answer
     
     

a) Select STAT, then 1:Edit and enter the data as shown in (a). Select 2nd STAT PLOT, then turn turn on Plot1 as shown in (b). Select scatterplot, choose the lists you used for the data, then a marker, as shown in (b). Finally, select 9:ZoomStat from the ZOOM menu to produce the plot in (c).

     

Screen Shot 2019-10-02 at 4.11.58 PM.png

     

b) Push the STAT button, right-arrow to the CALC menu, then select 4:LinReg(ax+b), followed by L1, a comma, L2, a comma, and Y1 from the VARS menu (select Y-VARS then 1:Function submenus). The result is shown in (a). Press ENTER to produce the calculation shown in (b). This is the line of best fit. This procedure also stores the line of best fit in Y1, so pushing the GRAPH button produces the line of best fit shown in (c).

     

Screen Shot 2019-10-02 at 4.12.45 PM.png

     

In (b) above, we see that the number sold N, as a function of average price P is given by the linear function N(P) = 0.24P − 40.33.

     

c) If all went well in part (b), then we can press 2nd TBL SET and set up the table parameters as shown in (g). Make sure the independent variable is set to ASK. Select 2nd TABLE, then enter 550 (which represents an average price P = $550, 000) as shown in (h).

     

Screen Shot 2019-10-02 at 4.14.24 PM.png

     

Thus, N(550) = 91.667, so approximately 91.667 thousand or 91, 667 homes are sold at the average price of $550,000. d) Press the STAT button, select 1:Edit and add the point (550, 41) to the table, as shown in (i). Select 9:ZoomStat from the ZOOM menu to produce the image in (j). Note that the data no longer depicts a linear trend, but something more of a nonlinear (curvy) nature

     

Screen Shot 2019-10-02 at 4.15.21 PM.png

     

e) In (k) we’ve drawn a smooth curve of best fit.

     

Screen Shot 2019-10-02 at 4.16.07 PM.png

     
 
 
 
 

Exercise (PageIndex{10})

 

The following from the July 14, 2006 edition of the Beijing Today newspaper shows how high-heels affect the ball of the foot. The table shows the increase in percent of pressure on the ball of the foot for given heights of heels.

                                                                                                                                                                           
Heel height h (inches) %increase in pressure
1 22
2 57
 

a) Enter the data into your calculator and make a scatter plot. Copy it down onto your paper, labeling appropriately.

 

b) Notice that, because we have exactly two data points, the line of best fit is the line that goes through both points. To begin finding the equation, use the slope formula to compute the slope.

 

c) Use the point-slope form to find an equation for the line. Write it in slope-intercept form.

 

d) Use the linear function to predict the percent of stress increase for a 3-inch heel.

 

e) The actual percent of pressure increase for a 3-inch heel is 76 %. Plot this as a point on your coordinate system and compare it to your linear function model’s prediction. Notice that this actual value is pretty different from the prediction.

 

f) What this means is that a linear model is not very good for the data! Draw a simple curve that goes through each of the data points. Notice that it does not very closely resemble the shape of a line! More sophisticated functions are required to model this example. Not every data set should be modeled linearly!

 
 

 

 

 

 
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