2.1 Ejercicios
En los Ejercicios ( PageIndex {1} ) – ( PageIndex {6} ), indique el dominio y el rango de la relación dada.
Ejercicio ( PageIndex {1} )
R = {(1, 3), (2, 4), (3, 4)}
- Respuesta
-
El dominio es el conjunto de todas las primeras coordenadas = {1, 2, 3}. El rango es el conjunto de todas las segundas coordenadas {3, 4} (tenga en cuenta que en un conjunto no enumera un objeto dos veces, por lo que solo enumeramos 4 una vez).
Ejercicio ( PageIndex {2} )
R = {(1, 3), (2, 4), (2, 5)}
Ejercicio ( PageIndex {3} )
R = {(1, 4), (2, 5), (2, 6)}
- Respuesta
-
El dominio es el conjunto de todas las primeras coordenadas = {1, 2} (tenga en cuenta que en un conjunto no enumera un objeto dos veces, por lo que solo enumeramos 2 una vez). El rango es el conjunto de todas las segundas coordenadas {4, 5, 6}.
Ejercicio ( PageIndex {4} )
R = {(1, 5), (2, 4), (3, 6)}
Ejercicio ( PageIndex {5} )
- Respuesta
-
Lea la coordenada x de cada punto para obtener que el dominio es {1, 2, 3}. Luego lea las coordenadas y para obtener que el rango es {1, 2, 3, 4}
Ejercicio ( PageIndex {6} )
En los Ejercicios ( PageIndex {7} ) – ( PageIndex {12} ), cree un diagrama de mapeo para la relación dada y establezca si es o no una función.
Ejercicio ( PageIndex {7} )
La relación en el ejercicio ( PageIndex {1} ).
- Respuesta
-
Crear un diagrama de mapeo para R.
Dado que ningún valor de dominio está emparejado con dos valores de rango, esta es una función (cada x se asigna a una sola y). Tenga en cuenta que tener dos valores de dominio diferentes van a un solo valor de rango (2 y 3 se asignan a 4) es permisible para una función.
Ejercicio ( PageIndex {8} )
La relación en el ejercicio ( PageIndex {2} ).
Ejercicio ( PageIndex {9} )
La relación en el ejercicio ( PageIndex {3} ).
- Respuesta
-
Crear un diagrama de mapeo para R.
El número 2 se asigna a dos valores de rango diferentes (uno x se asigna a dos y), por lo que esta no es una función.
Ejercicio ( PageIndex {10} )
La relación en el ejercicio ( PageIndex {4} ).
Ejercicio ( PageIndex {11} )
La relación en el ejercicio ( PageIndex {5} ).
- Respuesta
-
Crear un diagrama de mapeo para R.
El número 3 se asigna a dos valores de rango diferentes (uno x se asigna a dos y), por lo que esta no es una función.
Ejercicio ( PageIndex {12} )
La relación en el ejercicio ( PageIndex {6} ).
Ejercicio ( PageIndex {13} )
Dado que g toma un número real y lo duplica, entonces (g: x rightarrow? ).
- Respuesta
-
Dobles significa «multiplica por 2», entonces (g: x rightarrow 2x ).
Ejercicio ( PageIndex {14} )
Dado que f toma un número real y lo divide por 3, entonces (f: x rightarrow? ).
Ejercicio ( PageIndex {15} )
Dado que g toma un número real y le agrega 3, entonces (g: x rightarrow? ).
- Respuesta
-
(g: x rightarrow x + 3 )
Ejercicio ( PageIndex {16} )
Dado que h toma un número real y resta 4 de él, entonces (h: x rightarrow? ).
Ejercicio ( PageIndex {17} )
Dado que g toma un número real, lo duplica, luego suma 5, luego (g: x rightarrow? )
- Respuesta
-
Para una x puesta en g, g la duplica, dando 2x, y luego suma cinco, resultando en 2x + 5. Por lo tanto, (g: x rightarrow 2x + 5 )
Ejercicio ( PageIndex {18} )
Dado que h toma un número real, resta 3 de él, luego divide el resultado por 4, luego (h: x rightarrow? )
Dado que la función f está definida por la regla (h: x rightarrow 3x – 5 ), determine dónde se asigna el número de entrada en los Ejercicios ( PageIndex {19} ) – ( PageIndex { 22} ).
Ejercicio ( PageIndex {19} )
(f: 3 rightarrow? )
- Respuesta
-
Pon 3 en f. Esto significa, reemplazar x con 3 y calcular la salida. (f: 3 rightarrow 3 (3) – 5 = 4 ), entonces (f: 3 rightarrow 4 ).
Ejercicio ( PageIndex {20} )
(f: -5 rightarrow? )
Ejercicio ( PageIndex {21} )
(f: a rightarrow? )
- Respuesta
-
Pon un a en f, como lo harías con un número. Esto significa, reemplazar x con a y calcular la salida. (f: a rightarrow 3 (a) – 5 = 3a – 5 ), entonces (f: 3 rightarrow 3a – 5 ).
Ejercicio ( PageIndex {22} )
(f: 2a + 3 rightarrow? )
Dado que la función f está definida por la regla (f: x rightarrow 4-5x ), determine dónde se asigna el número de entrada en los Ejercicios ( PageIndex {23} ) – ( PageIndex { 26} ).
Ejercicio ( PageIndex {23} )
(f: 2 rightarrow? )
- Respuesta
-
Ponga 2 en f reemplazando x con él. (f: 2 rightarrow 4 – 5 (2) = −6 ), entonces (f: 2 rightarrow −6 ).
Ejercicio ( PageIndex {24} )
(f: -3 rightarrow? )
Ejercicio ( PageIndex {25} )
(f: a rightarrow? )
- Respuesta
-
Pon una a en f reemplazando x con ella, tal como lo harías con un número. (f: a rightarrow 4 – 5 (a) ), entonces (f: 2 rightarrow 4 – 5a ).
Ejercicio ( PageIndex {26} )
(f: 2a + 11 rightarrow? )
Dado que la función f está definida por la regla (f: x rightarrow x ^ {2} -4x-6 ), determine dónde se asigna el número de entrada en los Ejercicios ( PageIndex {27} ) – ( PageIndex {30} ).
Ejercicio ( PageIndex {27} )
(f: 1 rightarrow? )
- Respuesta
-
Ponga 1 en f reemplazando x con él. (f: 1 rightarrow (1) 2 – 4 (1) – 6 = 1 – 4 – 6 = −9 ), entonces (f: 1 rightarrow −9 ).
Ejercicio ( PageIndex {28} )
(f: -2 rightarrow? )
Ejercicio ( PageIndex {29} )
(f: -1 rightarrow? )
- Respuesta
-
Ponga −1 en f reemplazando x con él. (f: −1 rightarrow (−1) 2 −4 (−1) −6 = 1 + 4−6 = −1 ), entonces (f: 1 rightarrow −1 ).
Ejercicio ( PageIndex {30} )
(f: a rightarrow? )
Dado que la función f está definida por la regla (f: x rightarrow 3x-9 ), determine dónde se asigna el número de entrada en los Ejercicios ( PageIndex {31} ) – ( PageIndex { 34} ).
Ejercicio ( PageIndex {31} )
(f: a rightarrow? )
- Respuesta
-
Pon una a en f reemplazando x con ella, tal como lo harías con un número. (f: a rightarrow 3a – 9 ).
Ejercicio ( PageIndex {32} )
(f: a + 1 rightarrow? )
Ejercicio ( PageIndex {33} )
(f: 2a-5 rightarrow? )
- Respuesta
-
Pon 2a – 5 en f reemplazando x con él, tal como lo harías con un número. Obtenemos (f: 2a – 5 rightarrow 3 (2a – 5) – 9 = 6a – 15 – 9 = 6a – 24 ), entonces (f: 2a – 5 rightarrow 6a – 24 )
Ejercicio ( PageIndex {34} )
(f: a + h rightarrow? )
Dado que las funciones fyg están definidas por las reglas (f: x rightarrow 2x + 3 ) y (g: x rightarrow 4-x ), determine dónde se asigna el número de entrada en los Ejercicios ( PageIndex {35} ) – ( PageIndex {38} ).
Ejercicio ( PageIndex {35} )
(f: 2 rightarrow? )
- Respuesta
-
Ponga 2 en f reemplazando x con él. Obtenemos (f: 2 rightarrow 2 (2) +3 = 7 ), entonces (f: 2 rightarrow 7 ).
Ejercicio ( PageIndex {36} )
(f: 2 rightarrow? )
Ejercicio ( PageIndex {37} )
(f: a + 1 rightarrow? )
- Respuesta
-
Ponga a + 1 en f reemplazando x con él, tal como lo haría con un número. Obtenemos (f: a + 1 rightarrow 2 (a + 1) + 3 = 2a + 2 + 3 = 2a + 5 ), entonces (f: a + 1 rightarrow 2a + 5 )
Ejercicio ( PageIndex {38} )
(f: a-3 rightarrow? )
Ejercicio ( PageIndex {39} )
Dado que g toma un número real y lo triplica, entonces g (x) =?.
- Respuesta
-
Triples significa «multiplica por 3», entonces g (x) = 3x
Ejercicio ( PageIndex {40} )
Dado que f toma un número real y lo divide por 5, entonces f (x) =?.
Ejercicio ( PageIndex {41} )
Dado que g toma un número real y lo resta de 10, entonces g (x) =?.
- Respuesta
-
g toma una entrada x y la resta DE 10, entonces g (x) = 10 – x.
Ejercicio ( PageIndex {42} )
Dado que f toma un número real, lo multiplica por 5 y luego agrega 4 al resultado, luego f (x) =?.
Ejercicio ( PageIndex {43} )
Dado que f toma un número real, lo duplica, luego resta el resultado de 11, luego f (x) =?.
- Respuesta
-
f toma una entrada x, la duplica para obtener 2x, y la quita de 11, obteniendo 11 – 2x. Por lo tanto, f (x) = 11 – 2x.
Ejercicio ( PageIndex {44} )
Dado que h toma un número real, lo duplica, suma 5, luego toma la raíz cuadrada del resultado, luego h (x) =?.
En los Ejercicios ( PageIndex {45} ) – ( PageIndex {54} ), evalúa la función dada en el valor dado b.
Ejercicio ( PageIndex {45} )
f (x) = 12x + 2 para b = 6.
- Respuesta
-
Sustituye x por 6 en 12x + 2 y simplifica para obtener 74: f (6) = 12 (6) + 2 = 74.
Ejercicio ( PageIndex {46} )
f (x) = −11x – 4 para b = −3.
Ejercicio ( PageIndex {47} )
f (x) = −9x – 1 para b = −5.
- Respuesta
-
Sustituye −5 por x en −9x − 1 y simplifica para obtener 44: f (−5) = −9 (−5) −1 = 44.
Ejercicio ( PageIndex {48} )
f (x) = 11x + 4 para b = −4.
Ejercicio ( PageIndex {49} )
f (x) = 4 para b = −12.
- Respuesta
-
f es una función constante, entonces f (x) = 4 para todo x. Por lo tanto, f (−12) = 4.
Ejercicio ( PageIndex {50} )
f (x) = 7 para b = −7.
Ejercicio ( PageIndex {51} )
f (x) = 0 para b = −7.
- Respuesta
-
f es una función constante, entonces f (x) = 0 para todo x. Por lo tanto, f (−7) = 0.
Ejercicio ( PageIndex {52} )
f (x) = 12x + 8 para b = −3.
Ejercicio ( PageIndex {53} )
f (x) = −9x + 3 para b = −1.
- Respuesta
-
Sustituye −1 por x en −9x + 3 y simplifica para obtener 12: f (−1) = −9 (−1) +3 = 12
Ejercicio ( PageIndex {54} )
f (x) = 6x – 3 para b = 3.
En los ejercicios ( PageIndex {55} ) – ( PageIndex {58} ), dado que la función f está definida por la regla f (x) = 2x + 7, determine dónde se asigna el número de entrada .
Ejercicio ( PageIndex {55} )
f (a) =?
- Respuesta
-
Pon una a en f reemplazando x con ella, tal como lo harías con un número. Esto produce f (a) = 2a + 7.
Ejercicio ( PageIndex {56} )
f (a + 1) =?
Ejercicio ( PageIndex {57} )
f (3a – 2) =?
- Respuesta
-
Ponga 3a – 2 en f reemplazando x con él, tal como lo haría con un número. Esto produce f (3a – 2) = 2 (3a – 2) + 7 = 6a – 4 + 7 = 6a + 3.
Ejercicio ( PageIndex {58} )
f (a + h) =?
En los Ejercicios ( PageIndex {59} ) – ( PageIndex {62} ), dado que la función g está definida por la regla g (x) = 3 – 2x, determine dónde está el número de entrada mapeado.
Ejercicio ( PageIndex {59} )
g (a) =?
- Respuesta
-
Pon una a en g reemplazando x con ella, tal como lo harías con un número. Esto produce g (a) = 3 – 2a.
Ejercicio ( PageIndex {60} )
g (a + 3) =?
Ejercicio ( PageIndex {61} )
g (2 – 5a) =?
- Respuesta
-
Ponga 2 – 5a en g reemplazando x con él, tal como lo haría con un número. Esto produce g (2 – 5a) = 3 – 2 (2 – 5a) = 3 – 4 + 10a = −1 + 10a o 10a – 1.
Ejercicio ( PageIndex {62} )
g (a + h) =?
Dado que las funciones f y g están definidas por las reglas f (x) = 1 – x y g (x) = 2x + 13, determine dónde se asigna el número de entrada en los Ejercicios ( PageIndex {63} ) – ( PageIndex {66} ).
Ejercicio ( PageIndex {63} )
f (a) =?
- Respuesta
-
Pon una a en f reemplazando x con ella, tal como lo harías con un número. Esto produce f (a) = 1 – a.
Ejercicio ( PageIndex {64} )
g (a) =?
Ejercicio ( PageIndex {65} )
f (a + 3) =?
- Respuesta
-
Pon un a + 3 en f reemplazando x con él, tal como lo harías con un número. Esto produce f (a + 3) = 1 – (a + 3) = 1 – a – 3 = −a – 2.
Ejercicio ( PageIndex {66} )
g (4 – a) =?
Dado que las funciones f y g están definidas por las reglas f (x) = 3x + 4 y g (x) = 2x − 5, determine dónde se asigna el número de entrada en los Ejercicios ( PageIndex {67} ) – ( PageIndex {70} ).
Ejercicio ( PageIndex {67} )
f (g (2)) =?
- Respuesta
-
Primer cálculo g (2) = 2 (2) – 5 = −1. Esto significa que f (g (2)) es realmente f (−1). Al conectar −1 para x en la función f, obtenemos f (g (2)) = f (−1) = 3 (−1) + 4 = −3 + 4 = 1.
Ejercicio ( PageIndex {68} )
g (f (2)) =?
Ejercicio ( PageIndex {69} )
f (g (a)) =?
- Respuesta
-
Primer cálculo g (a) = 2a − 5. Esto significa que f (g (a)) es realmente f (2a − 5). Al conectar 2a – 5 para x en la función f, obtenemos f (g (a)) = f (2a – 5) = 3 (2a – 5) + 4 = 6a – 15 + 4 = 6a – 11. [19459006 ]
Ejercicio ( PageIndex {70} )
g (f (a)) =?
Dado que las funciones f y g están definidas por las reglas f (x) = 2x – 9 y g (x) = 11, determine dónde se asigna el número de entrada en los Ejercicios ( PageIndex {71} ) – ( PageIndex {74} ).
Ejercicio ( PageIndex {71} )
f (g (2)) =?
- Respuesta
-
Primero calcula g (2) = 11 (ten en cuenta que, no importa lo que pongas en g, genera 11). Esto significa que f (g (2)) es realmente f (11). Al conectar 11 para x en la función f, obtenemos f (g (2)) = f (11) = 2 (11) – 9 = 22 – 9 = 13.
Ejercicio ( PageIndex {72} )
g (f (2)) =?
Ejercicio ( PageIndex {73} )
f (g (a)) =?
- Respuesta
-
Primero calcula g (a) = 11 (ten en cuenta que, no importa lo que pongas en g, genera 11). Esto significa que f (g (a)) es realmente f (11). Al conectar 11 para x en la función f, obtenemos f (g (2)) = f (11) = 2 (11) – 9 = 22 – 9 = 13.
Ejercicio ( PageIndex {74} )
g (f (a)) =?
Utilice la notación de generador de conjuntos para describir el dominio de cada una de las funciones definidas en los Ejercicios ( PageIndex {75} ) – ( PageIndex {78} ).
Ejercicio ( PageIndex {75} )
(f (x) = dfrac {93} {x + 98} )
- Respuesta
-
Una entrada de x = −98 causaría la división por cero, por lo que −98 no está en el dominio. Todas las demás entradas posibles son válidas. El dominio, en notación de generador de conjuntos, es ( {x: x neq −98 } ).
Ejercicio ( PageIndex {76} )
(f (x) = dfrac {54} {x + 65} )
Ejercicio ( PageIndex {77} )
(f (x) = – dfrac {87} {x-88} )
- Respuesta
-
Una entrada de x = 88 causaría la división por cero, por lo que 88 no está en el dominio. Todas las demás entradas posibles son válidas. El dominio, en notación de generador de conjuntos, es ( {x: x neq 88 } ) ..
Ejercicio ( PageIndex {78} )
(f (x) = – dfrac {30} {x-52} )
Utilice el generador de conjuntos y la notación de intervalo para describir el dominio de las funciones definidas en los Ejercicios ( PageIndex {79} ) – ( PageIndex {82} ).
Ejercicio ( PageIndex {79} )
(f (x) = sqrt {x + 69} )
- Respuesta
-
La raíz cuadrada de un número negativo no se define como un número real. Por lo tanto, x + 69 debe ser mayor o igual que cero. Entonces (x + 69 geq 0 ) implica que (x geq −69 ), entonces el dominio es el intervalo ([- 69, infty) ), o en notación de generador de conjuntos, ( {x: x geq −69 } ).
Ejercicio ( PageIndex {80} )
(f (x) = sqrt {x + 62} )
Ejercicio ( PageIndex {81} )
(f (x) = sqrt {x-81} )
- Respuesta
-
La raíz cuadrada de un número negativo no se define como un número real. Por lo tanto, x – 81 debe ser mayor o igual que cero. Entonces (x – 81 geq 0 ) implica que (x geq 81 ), entonces el dominio es el intervalo ([81, infty) ), o en notación de generador de conjuntos, ( { x: x geq 81 } ).
Ejercicio ( PageIndex {82} )
(f (x) = sqrt {x-98} )
Se dice que dos enteros son relativamente primos si su máximo común divisor es 1. Por ejemplo, el máximo común divisor de 6 y 35 es 1, entonces 6 y 35 son relativamente primos. Por otro lado, el máximo común divisor de 14 y 21 no es 1 (es 7), por lo que 14 y 21 no son relativamente primos. La función Euler ( phi ) se define de la siguiente manera:
• Si n = 1, entonces ( phi (n) = 1 ).
• Si n> 1, entonces ( phi (n) ) es el número de enteros positivos menores que n que son relativamente primos para n. En los Ejercicios ( PageIndex {83} ) – ( PageIndex {84} ), evalúe la función Euler ( phi ) – en la entrada dada.
Ejercicio ( PageIndex {83} )
( phi (12) )
- Respuesta
-
1, 5, 7 y 11 son menores que 12 y cada uno es relativamente primo para 12. Por lo tanto, ( phi (12) = 4 ).
Ejercicio ( PageIndex {84} )
( phi (36) )
2.2 Ejercicios
Realice cada una de las siguientes tareas para las funciones definidas por las ecuaciones en los Ejercicios ( PageIndex {1} ) – ( PageIndex {8} ).
i. Establezca una tabla de puntos que satisfaga la ecuación dada. Coloque esta tabla de puntos al lado de su gráfico en su papel cuadriculado.
ii. Configure un sistema de coordenadas en una hoja de papel cuadriculado. Rotule y escale cada eje, luego trace cada uno de los puntos de su tabla en su sistema de coordenadas.
iii. Si está seguro de que «ve» la forma del gráfico, haga un «salto de fe» y trace todos los pares que satisfagan la ecuación dada dibujando una curva suave (a mano libre) en su sistema de coordenadas que contiene todos los gráficos previamente graficados puntos (use una regla solo si la gráfica de la ecuación es una línea). Si no está seguro de que «ve» la forma del gráfico, agregue más puntos a su tabla, grábelos en su sistema de coordenadas y vea si esto ayuda. Continúe este proceso hasta que «vea» la forma del gráfico y pueda completar el resto de los puntos que satisfacen la ecuación dibujando una curva suave (o línea) en su sistema de coordenadas.
Ejercicio ( PageIndex {1} )
(f (x) = 2x + 1 )
- Respuesta
-
Evalúa la función (f (x) = 2x + 1 ) en −2, −1, 0 y 1.
[ begin {array} {ccc} f (−2) & = & 2 (−2) + 1 & = & −3 \ f (−1) & = & 2 (−1) + 1 & = & −1 \ f (0) & = & 2 (0) + 1 & = & 1 \ f (1) & = & 2 (1) + 1 & = & 3 end {array} ] [ 19459006]
Coloque estos resultados en la tabla (a) y grábelos como se muestra en (b). Aquí hay suficiente evidencia para intuir que la gráfica de f es la línea que se muestra en (b).
x (f (x) = 2x + 1 ) (x, f (x)) -2 -3 (−2, −3) -1 -1 (−1, −1) 0 1 (0,1) 1 3 (1,3) - (a)
-
Ejercicio ( PageIndex {2} )
(f (x) = 1 – x )
Ejercicio ( PageIndex {3} )
(f (x) = 3 – dfrac {1} {2} x )
- Respuesta
-
Evalúe la función f (x) = 3 – (1/2) x en x = −2, 0, 2 y 4.
[f (−2) = 3 – (1/2) (- 2) = 4 \ f (0) = 3 – (1/2) (0) = 3 \ f (2) = 3 – (1/2) (2) = 2 \ f (4) = 3 – (1/2) (4) = 1 ]
Coloque estos resultados en la tabla (a) y grábelos como se muestra en (b). Aquí hay suficiente evidencia para intuir que la gráfica de f es la línea que se muestra en (b).
x (f (x) = 3 – x / 2 ) (x, f (x)) -2 4 (−2, 4) 0 3 (0,3) 2 2 (2,2) 4 1 (4,1)
(a)
Ejercicio ( PageIndex {4} )
(f (x) = −1 + dfrac {1} {2} x )
Ejercicio ( PageIndex {5} )
(f (x) = x ^ 2 – 2 )
- Respuesta
-
Evalúe (f (x) = x ^ 2 – 2 ) en x = −3, −2, −1, 0, 1, 2 y 3.
[f (−3) = (−3) ^ 2 – 2 = 7 \ f (−2) = (−2) ^ 2 – 2 = 2 \ f (−1) = (−1 ) ^ 2 – 2 = −1 \ f (0) = (0) ^ 2 – 2 = −2 \ f (1) = (1) ^ 2 – 2 = −1 \ f (2) = ( 2) ^ 2 – 2 = 2 \ f (3) = (3) ^ 2 – 2 = 7 ]
Coloque estos resultados en la tabla (a) y grábelos como se muestra en (b). Aquí hay suficiente evidencia para intuir que la gráfica de f es la curva que se muestra en (b).
x (f (x) = x ^ 2 – 2 ) (x, f (x)) -3 7 (−3, 7) -2 2 (-2,2) -1 -1 (-1, -1) 0 -2 (0, -2) 1 -1 (1, -1) 2 2 (2,2) 3 7 (3,7) (a)
Ejercicio ( PageIndex {6} )
(f (x) = 4 – x ^ {2} )
Ejercicio ( PageIndex {7} )
(f (x) = dfrac {1} {2} x ^ {2} – 6 )
- Respuesta
-
Evalúe (f (x) = x ^ 2/2 – 6 ) en x = −4, −2, 0, 2 y 4.
[f (−4) = (−4) ^ 2/2 – 6 = 2 \ f (−2) = (−2) ^ 2/2 – 6 = −4 \ f (0) = (0) ^ 2/2 – 6 = −6 \ f (2) = (2) ^ 2/2 – 6 = −4 \ f (4) = (4) ^ 2/2 – 6 = 2 ]
Coloque estos resultados en la tabla (a) y grábelos como se muestra en (b). Aquí hay suficiente evidencia para intuir que la gráfica de f es la curva que se muestra en (b).
x (f (x) = x ^ 2 – 2 ) (x, f (x)) -4 2 (−4, 2) -2 -4 (-2, -4) 0 -6 (0, -6) 2 -4 (2, -4) 4 2 (4,2)
(a)
Ejercicio ( PageIndex {8} )
(f (x) = 8- dfrac {1} {2} x ^ 2 )
Realice cada una de las siguientes tareas para las funciones Ejercicios ( PageIndex {9} ) – ( PageIndex {10} ).
i. Configure un sistema de coordenadas en una hoja de papel cuadriculado. Rotula y escala cada eje.
ii. Use la función de tabla de su calculadora gráfica para evaluar la función con los valores dados de x. Registre estos resultados en una tabla al lado de su sistema de coordenadas en su papel cuadriculado.
iii. Trace los puntos en la tabla en su sistema de coordenadas y luego úselos para dibujar la gráfica de la función dada. Rotula la gráfica con su ecuación.
Ejercicio ( PageIndex {9} )
(f (x) = sqrt {x – 4} ) en x = 4, 5, 6, 7, 8, 9 y 10.
- Respuesta
-
Cargue la función (f (x) = sqrt {x – 4} ) en Y1 como se muestra en (a). Seleccione TBLSET, luego resalte ASK para la variable independiente y presione ENTER (vea (b)). No importa lo que se ingrese para TblStart o ∆ Tbl. Seleccione TABLE e ingrese los valores x 4, 5, 6, 7, 8, 9 y 10, como se muestra en (c).
Trace los puntos en la tabla (c) en (d). Esto es suficiente para intuir que la gráfica de f es la curva que se muestra en (d).
Ejercicio ( PageIndex {10} )
(f (x) = sqrt {4 – x} ) en x = −10, −8, −6, −4, −2, 0, 2 y 4.
En los Ejercicios ( PageIndex {11} ) – ( PageIndex {14} ), la gráfica de la función dada es una parábola, una gráfica que tiene una «forma de U». Una parábola tiene solo un punto de inflexión. Para cada ejercicio, realice las siguientes tareas.
i. Cargue la ecuación en el menú Y = de su calculadora gráfica. Ajuste los parámetros de VENTANA para que el «punto de inflexión» (en realidad llamado vértice) sea visible en la ventana de visualización.
ii. Haga una copia razonable de la imagen en la ventana de visualización en su tarea. Dibuje todas las líneas con una regla (incluidos los ejes), pero dibuje curvas a mano alzada. Rotule y escale cada eje con xmin, xmax, ymin e ymax. Rotula la gráfica con su ecuación.
Ejercicio ( PageIndex {11} )
(f (x) = x ^ {2} – x – 30 )
- Respuesta
-
Cargue la función (f (x) = x ^ {2} – x – 30 ) en Y1 como se muestra en (a). Ajuste los parámetros de la VENTANA como se muestra en (b). Presione el botón GRAPH para obtener la gráfica de f en (c).
Copie la imagen en su tarea como se muestra en (d).
Ejercicio ( PageIndex {12} )
(f (x) = 24 – 2x – x ^ 2 )
Ejercicio ( PageIndex {13} )
(f (x) = 11 + 10x – x ^ 2 )
- Respuesta
-
Cargue la función (f (x) = 11 + 10x – x ^ 2 ) en Y1 como se muestra en (a). Ajuste los parámetros de la VENTANA como se muestra en (b). Presione el botón GRAPH para obtener la gráfica de f en (c).
Copie la imagen en su tarea como se muestra en (d).
Ejercicio ( PageIndex {14} )
(f (x) = x ^ 2 + 11x – 12 )
Cada una de las ecuaciones en los Ejercicios ( PageIndex {15} ) – ( PageIndex {18} ) se llaman «polinomios cúbicos». Cada ecuación ha sido cuidadosamente elegida para que su gráfico tenga exactamente dos «puntos de inflexión». Para cada ejercicio, realice cada una de las siguientes tareas. yo. Cargue la ecuación en el menú Y = de su calculadora gráfica y ajuste los parámetros de VENTANA para que ambos «puntos de giro» sean visibles en la ventana de visualización. ii) Haga una copia razonable del gráfico en la ventana de visualización de su tarea. Rotula y escala cada eje con xmin, xmax, ymin e ymax, luego rotula la gráfica con su ecuación. Recuerde dibujar todas las líneas con una regla.
Ejercicio ( PageIndex {15} )
(f (x) = x ^ 3 – 2x ^ 2 – 29x + 30 )
- Respuesta
-
Cargue la función (f (x) = x ^ 3 – 2x ^ 2 – 29x + 30 ) en Y1 como se muestra en (a). Ajuste los parámetros de la VENTANA como se muestra en (b). Presione el botón GRAPH para obtener la gráfica de f en (c).
Copie la imagen en su tarea como se muestra en (d).
Exercise (PageIndex{16})
(f(x) = −x^3 + 2x^2 + 19x − 20)
Exercise (PageIndex{17})
(f(x) = x^3 + 8x^2 − 53x − 60)
- Answer
-
Load the function (f(x) = x^3 + 8x^2 − 53x − 60) into Y1 as shown in (a). Adjust the WINDOW parameters as shown in (b). Push the GRAPH button to obtain the graph of f in (c).
Copy the image onto your homework as shown in (d).
Exercise (PageIndex{18})
(f(x) = −x^3 + 16x^2 − 43x − 60)
Perform each of the following tasks for the equations in Exercises (PageIndex{19})-(PageIndex{22}).
i. Load the equation into the Y= menu. Adjust the WINDOW parameters until you think all important behavior (“turning points,” etc.) is visible in the viewing window. Note: This is more difficult than it sounds, particularly when we have no advance notion of what the graph might look like. However, experiment with several settings until you “discover” the settings that exhibit the most important behavior.
ii. Copy the image on the screen onto your homework paper. Label and scale each axis with xmin, xmax, ymin, and ymax. Rotula la gráfica con su ecuación.
Exercise (PageIndex{19})
(f(x) = 2x^2 − x − 465)
- Answer
-
Load the function (f(x) = 2x^2 − x − 465) into Y1 as shown in (a). Adjust the WINDOW parameters as shown in (b). Push the GRAPH button to obtain the graph of f in (c).
Copy the image onto your homework as shown in (d).
Exercise (PageIndex{20})
(f(x) = x^3 − 24x^2 + 65x + 1050)
Exercise (PageIndex{21})
(f(x) = x^4 − 2x^3 − 168x^2 + 288x + 3456)
- Answer
-
Load the function (f(x) = x^4 − 2x^3 − 168x^2 + 288x + 3456) into Y1 as shown in (a). Adjust the WINDOW parameters as shown in (b). Push the GRAPH button to obtain the graph of f in (c).
Copy the image onto your homework as shown in (d)
Exercise (PageIndex{22})
(f(x) = −x^4 −3x^3 +141x^2 +523x− 660)
2.3 Exercises
For Exercises (PageIndex{1})-(PageIndex{6}), perform each of the following tasks.
i. Make a copy of the graph on a sheet of graph paper and apply the vertical line test.
ii. Write a complete sentence stating whether or not the graph represents a function. Explain the reason for your response.
Exercise (PageIndex{1})
- Answer
-
Note that in the figure below a vertical line cuts the graph more than once. Therefore, the graph does not represent the graph of a function.
Exercise (PageIndex{2})
Exercise (PageIndex{3})
- Answer
-
No vertical line cuts the graph more than once (see figure below). Therefore, the graph represents a function.
Exercise (PageIndex{4})
Exercise (PageIndex{5})
- Answer
-
Note that in the figure below a vertical line cuts the graph more than once. Therefore, the graph does not represent the graph of a function.
Exercise (PageIndex{6})
In Exercises (PageIndex{7})-(PageIndex{12}), perform each of the following tasks.
i. Make an exact copy of the graph of the function f on a sheet of graph paper. Label and scale each axis. Remember to draw all lines with a ruler.
ii. Use the technique of Examples 3 and 4 in the narrative to evaluate the function at the given value. Draw and label the arrows as shown in Figures 4 and 5 in the narrative.
Exercise (PageIndex{7})
Use the graph of f to determine f(2).
- Answer
-
Locate x = 2 on the x-axis (see figure below), draw a vertical arrow to the graph of f, then a horizontal arrow to the y-axis. Thus, f(2) = −1.
Exercise (PageIndex{8})
Use the graph of f to determine f(3).
Exercise (PageIndex{9})
Use the graph of f to determine f(−2).
- Answer
-
Locate x = −2 on the x-axis (see figure below), draw a vertical arrow to the graph of f, then a horizontal arrow to the y-axis. Thus, f(−2) = 1.
Exercise (PageIndex{10})
Use the graph of f to determine f(1).
Exercise (PageIndex{11})
Use the graph of f to determine f(1).
- Answer
-
Locate x = 1 on the x-axis (see figure below), draw a vertical arrow to the graph of f, then a horizontal arrow to the y-axis. Thus, f(1) = 3.
Exercise (PageIndex{12})
Use the graph of f to determine f(−2).
In Exercises (PageIndex{13})-(PageIndex{18}), perform each of the following tasks.
i. Make an exact copy of the graph of the function f on a sheet of graph paper. Label and scale each axis. Remember to draw all lines with a ruler.
ii. Use the technique of Example 5 in the narrative to find the value of x that maps onto the given value. Draw and label the arrows as shown in Figure 6 in the narrative.
Exercise (PageIndex{13})
Use the graph of f to solve the equation f(x) = −2.
- Answer
-
Locate y = −2 on the y-axis (see figure below), draw a horizontal arrow to the graph of f, then a vertical arrow to the y-axis. Thus, the solution of f(x) = −2 is x = −3.
Exercise (PageIndex{14})
Use the graph of f to solve the equation f(x) = 1.
Exercise (PageIndex{15})
Use the graph of f to solve the equation f(x) = 2
- Answer
-
Locate y = 2 on the y-axis (see figure below), draw a horizontal arrow to the graph of f, then a vertical arrow to the y-axis. Thus, the solution of f(x) = 2 is x = −2.
Exercise (PageIndex{16})
Use the graph of f to solve the equation f(x) = −2.
Exercise (PageIndex{17})
Use the graph of f to solve the equation f(x) = 2.
- Answer
-
Locate y = 2 on the y-axis (see figure below), draw a horizontal arrow to the graph of f, then a vertical arrow to the y-axis. Thus, the solution of f(x) = 2 is x = −1.
Exercise (PageIndex{18})
Use the graph of f to solve the equation f(x) = −3.
In the Exercises (PageIndex{19})-(PageIndex{22}), perform each of the following tasks.
i. Make a copy of the graph of f on a sheet of graph paper. Label and scale each axis.
ii. Using a different colored pen or pencil, project each point on the graph of f onto the x-axis. Shade the resulting domain on the x-axis.
iii. Use both set-builder and interval notation to describe the domain.
Exercise (PageIndex{19})
- Answer
-
To find the domain of the function, project the graph of f onto the x-axis. Note that all values of x that lie to the right of −3 lie in shadow and are hence in the domain of f. Therefore, the domain is best described with the notation ({x : x > −3} = (−3,infty)).
Exercise (PageIndex{20})
Exercise (PageIndex{21})
- Answer
-
To find the domain of the function, project the graph of f onto the x-axis. Note that all values of x that lie to the left of 0 lie in shadow and are hence in the domain of f. Therefore, the domain is best described with the interval notation ({x : x < 0} = (−infty, 0)).
Exercise (PageIndex{22})
In Exercises (PageIndex{23})-(PageIndex{26}), perform each of the following tasks.
i. Make a copy of the graph of f on a sheet of graph paper. Label and scale each axis.
ii. Using a different colored pen or pencil, project each point on the graph of f onto the y-axis. Shade the resulting range on the y-axis. iii. Use both set-builder and interval notation to describe the range.
Exercise (PageIndex{23})
- Answer
-
To find the range of the function, project the graph of f onto the y-axis. Note that all values of y that lie below 1 lie in shadow and are hence in the range of f. Therefore, the range is best described with the interval notation ({y : y < 1} = (−infty, 1)).
Exercise (PageIndex{24})
Exercise (PageIndex{25})
- Answer
-
To find the range of the function, project the graph of f onto the y-axis. Note that all values of y that lie above −2 lie in shadow and are hence in the range of f. Therefore, the range is best described with the interval notation ({y : y > −2} = (−2,infty)).
Exercise (PageIndex{26})
In Exercises (PageIndex{27})-(PageIndex{30}), perform each of the following tasks.
i. Use your graphing calculator to draw the graph of the given function. Make a reasonably accurate copy of the image in your viewing screen on your homework paper. Label and scale each axis with the WINDOW parameters xmin, xmax, ymin, and ymax. Rotula la gráfica con su ecuación.
ii. Using a colored pencil, project each point on the graph onto the x-axis; i.e., shade the domain on the x-axis. Use interval and set-builder notation to describe the domain.
iii. Use a purely algebraic technique, as demonstrated in Example 8 in the narrative, to find the domain. Compare this result with that found in part (ii).
iv. Using a different colored pencil, project each point on the graph onto the y-axis; i.e., shade the range on the y-axis. Use interval and set-builder notation to describe the range.
Exercise (PageIndex{27})
(f(x) = sqrt{x + 5}).
- Answer
-
Load the function (f(x) = sqrt{x + 5}) into Y1 as shown in (a). Select 6:ZStandrd from the ZOOM menu to produce the graph in (b).
Copy the image in (b) onto your homework paper, then project the domain and range onto the x- and y-axes, as shown in (c) and (d), respectively.
To find the domain algebraically, note that you cannot take the square root of a negative number, so the expression under the radical in (f(x) = sqrt{x + 5}), namely x+5, must either be positive or zero (nonnegative). That is,
[x + 5 geq 0]
or equivalently,
[xgeq -5]
Thus, the domain of f is Domain = ([−5,infty)), or in set-builder notation, Domain = ({x : x geq −5}).
Exercise (PageIndex{28})
(f(x) = sqrt{5-x})
Exercise (PageIndex{29})
(f(x) = − sqrt{4 − x}).
- Answer
-
Load the function (f(x) = − sqrt{4 − x}). into Y1 as shown in (a). Select 6:ZStandrd from the ZOOM menu to produce the graph in (b).
Copy the image in (b) onto your homework paper, then project the domain and range onto the x- and y-axes, as shown in (c) and (d), respectively.
To find the domain algebraically, note that you cannot take the square root of a negative number, so the expression under the radical in (f(x) = − sqrt{4 − x}), namely 4−x, must either be positive or zero (nonnegative). That is,
[4-xgeq 0]
or equivalently,
[-x geq -4 \ xleq 4]
Thus, the domain of f is Domain = ((infty, 4]), or in set-builder notation, Domain = ({x : x leq 4}).
Exercise (PageIndex{30})
(f(x) = − sqrt{x + 4})
2.4 Exercises
In Exercises (PageIndex{1})-(PageIndex{6}), you are given the definition of two functions f and g. Compare the functions, as in Example 1 of the narrative, at the given values of x.
Exercise (PageIndex{1})
f(x) = x+2, g(x) = 4−x at x = −3, 1, and 2.
- Answer
-
We’re given that f(x) = x + 2 and g(x) = 4 − x. At x = −3,
[f(−3) = −3 + 2 = −1 \ g(−3) = 4 − (−3) = 7]
Therefore, f(−3) < g(−3). At x = 1,
[f(1) = 1 + 2 = 3 \ g(1) = 4 − 1 = 3.]
Therefore, f(1) = g(1). At x = 2,
[f(2) = 2 + 2 = 4 \ g(2) = 4 − 2 = 2.]
Therefore, f(2) > g(2).
Exercise (PageIndex{2})
f(x) = 2x − 3, g(x) = 3 − x at x = −4, 2, and 5.
Exercise (PageIndex{3})
f(x) = 3−x, g(x) = x+9 at x = −4, −3, and −2.
- Answer
-
We’re given that f(x) = 3 − x and g(x) = x + 9. At x = −4,
[f(−4) = 3 − (−4) = 7 \ g(−4) = −4 + 9 = 5]
Therefore, f(−4) > g(−4). At x = −3,
[f(−3) = 3 − (−3) = 6 \ g(−3) = −3 + 9 = 6]
Therefore, f(−3) = g(−3). At x = −2,
[f(−2) = 3 − (−2) = 5 \ g(−2) = −2 + 9 = 7]
Therefore, f(−2) < g(−2).
Exercise (PageIndex{4})
(f(x) = x^2), g(x) = 4x + 5 at x = −2, 1, and 6.
Exercise (PageIndex{5})
(f(x) = x^2), g(x) = −3x − 2 at x = −3, −1, and 0.
- Answer
-
We’re given that (f(x) = x^2) and g(x) = −3x − 2. At x = −3,
[f(−3) = (−3)2 = 9 \ g(−3) = −3(−3) − 2 = 7]
Therefore, f(−3) > g(−3). At x = −1,
[f(−1) = (−1)2 = 1 \ g(−1) = −3(−1) − 2 = 1]
Therefore, f(−1) = g(−1). At x = 0,
[f(0) = (0)2 = 0 \ g(0) = −3(0) − 2 = −2]
Therefore, f(0) > g(0).
Exercise (PageIndex{6})
f(x) = |x|, g(x) = 4 − x at x = 1, 2, and 3.
In Exercises (PageIndex{7})-(PageIndex{12}), perform each of the following tasks. Remember to use a ruler to draw all lines.
i. Make an accurate copy of the image on graph paper (label each equation, label and scale each axis), drop a dashed vertical line through the point of intersection, then label and shade the solution of f(x) = g(x) on the x-axis.
ii. Make a second copy of the image on graph paper, drop a dashed, vertical line through the point of intersection, then label and shade the solution of f(x) > g(x) on the x-axis. Use set-builder and interval notation to describe your solution set.
iii. Make a third copy of the image on graph paper, drop a dashed, vertical line through the point of intersection, then label and shade the solution of f(x) < g(x) on the x-axis. Use setbuilder and interval notation to describe your solution set.
Exercise (PageIndex{7})
- Answer
-
The graph of f intersects the graph of g at x = 3. The solution of f(x) = g(x) is x = 3.
The graph of f lies above the graph of g to the right of x = 3. The solution of f(x) > g(x) is ((3,infty) = {x : x > 3}).
The graph of f lies below the graph of g to the left of x = 3. The solution of f(x) < g(x) is ((−infty, 3) = {x : x < 3}).
Exercise (PageIndex{8})
Exercise (PageIndex{9})
- Answer
-
The graph of f intersects the graph of g at x = −2. The solution of f(x) = g(x) is x = −2.
The graph of f lies above the graph of g to the left of x = −2. The solution of f(x) > g(x) is ((−infty, −2) = {x : x < −2}).
The graph of f lies below the graph of g to the right of x = −2. The solution of f(x) < g(x) is ((−2,infty) = {x : x > −2}).
Exercise (PageIndex{10})
Exercise (PageIndex{11})
- Answer
-
The graph of f intersects the graph of g at x = 3. The solution of f(x) = g(x) is x = 3.
The graph of f is above the graph of g to the right of x = 3. The solution of f(x) > g(x) is ((3,infty) = {x : x > 3}).
The graph of f is below the graph of g to the left of x = 3. The solution of f(x) < g(x) is ((−infty, 3) = {x : x < 3}).
Exercise (PageIndex{12})
In Exercises (PageIndex{13})-(PageIndex{16}), perform each of the following tasks. Remember to use a ruler to draw all lines.
i. Make an accurate copy of the image on graph paper, drop dashed, vertical lines through the points of intersection, then label and shade the solution of (f(x) geq g(x)) on the x-axis. Use set-builder and interval notation to describe your solution set.
ii. Make a second copy of the image on graph paper, drop dashed, vertical lines through the points of intersection, then label and shade the solution of f(x) < g(x) on the x-axis. Use set-builder and interval notation to describe your solution set.
Exercise (PageIndex{13})
- Answer
-
The graph of f intersects the graph of g at x = −3 and x = 3. The graph of f lies above the graph of g for values of x that lie between −3 and 3. Therefore, the solution of (f(x) geq g(x)) is ([−3, 3] = {x : −3 leq x leq 3}).
The graph of f is below the graph of g for values of x that lie to the left of −3 or to the right of 3. Therefore, the solution of f(x) < g(x) is ((−infty, −3) cup (3,infty)) or ({x : x < −3 or x > 3}).
Exercise (PageIndex{14})
Exercise (PageIndex{15})
- Answer
-
The graph of f intersects the graph of g at x = −2 and at x = 2. The graph of f lies above the graph of g for all values of x that lie to the left of −2 or to the right of 2. Therefore, the solution of (f(x) geq g(x)) is ((−infty, −2] cup [2,infty)) or ({x : x leq −2 or x geq 2}).
The graph of f lies below the graph of g for values of x that lie between −2 and 2. Therefore, the solution of f(x) < g(x) is ((−2, 2) = {x : −2 < x < 2}).
Exercise (PageIndex{16})
In Exercises (PageIndex{17})-(PageIndex{20}), perform each of the following tasks. Remember to use a ruler to draw all lines.
i. Load each side of the equation into the Y= menu of your calculator. Adjust the WINDOW parameters so that the point of intersection of the graphs is visible in the viewing window. Use the intersect utility in the CALC menu of your calculator to determine the x-coordinate of the point of intersection.
ii. Make an accurate copy of the image in your viewing window on your homework paper. Label and scale each axis with xmin, xmax, ymin, and ymax, and label each graph with its equation.
iii. Draw a dashed, vertical line through the point of intersection. Shade and label the solution of the equation on the x-axis.
Exercise (PageIndex{17})
1.23x − 4.56 = 3.46 − 2.3x
- Answer
-
To solve the equation 1.23x − 4.56 = 3.46 − 2.3x graphically, start by loading the left- and right-hand sides of the equation into Y1 and Y2, respectively, as shown in (a). Use the intersect utility in the CALC menu to determine the point of intersection, as shown in (c).
Therefore, the solution of the equation is x = 2.2719547, which is shaded on the x-axis in the image that follows. Answers may vary due to roundoff error.
Exercise (PageIndex{18})
2.23x − 1.56 = 5.46 − 3.3x
Exercise (PageIndex{19})
5.46 − 1.3x = 2.2x − 5.66
- Answer
-
To solve the equation 5.46 − 1.3x = 2.2x − 5.66 graphically, start by loading the left- and right-hand sides of the equation into Y1 and Y2, respectively, as shown in (a). Use the intersect utility in the CALC menu to determine the point of intersection, as shown in (c).
Therefore, the solution of the equation is x = 3.1771429, which is shaded on the x-axis in the image that follows. Answers may vary due to roundoff error.
Exercise (PageIndex{20})
2.46 − 1.4x = 1.2x − 2.66
In Exercises (PageIndex{21})-(PageIndex{26}), perform each of the following tasks. Remember to use a ruler to draw all lines.
i. Load each side of the inequality into the Y= menu of your calculator. Adjust the WINDOW parameters so that the point(s) of intersection of the graphs is visible in the viewing window. Use the intersect utility in the CALC menu of your calculator to determine the coordinates of the point(s) of intersection.
ii. Make an accurate copy of the image in your viewing window on your homework paper. Label and scale each axis with xmin, xmax, ymin, and ymax, and label each graph with its equation.
iii. Draw a dashed, vertical line through the point(s) of intersection. Shade and label the solution of the inequality on the x-axis. Use both set-builder and interval notation to describe the solution set.
Exercise (PageIndex{21})
(1.6x + 1.23 geq −2.3x − 4.2)
- Answer
-
To solve the inequality (1.6x+1.23 geq −2.3x−4.2) graphically, start by loading the left- and right-hand sides of the inequality into Y1 and Y2, respectively, as shown in (a). Use the intersect utility in the CALC menu to determine the point of intersection, as shown in (c).
The two graphs intersect at x = −1.392308. The graph of y = 1.6x + 1.23 is above the graph of y = −2.3x−4.2 for all values of x that lie to the right of −1.392308. Therefore, the solution of (1.6x + 1.23 geq −2.3x − 4.2) is ([−1.392308,infty) = {x : x geq −1.392308}).
Exercise (PageIndex{22})
1.24x + 5.6 < 1.2 − 0.52x
Exercise (PageIndex{23})
0.15x − 0.23 > 8.2 − 0.6x
- Answer
-
To solve the inequality 0.15x − 0.23 > 8.2 − 0.6x graphically, start by loading the left- and right-hand sides of the inequality into Y1 and Y2, respectively, as shown in (a). Adjust the viewing window as shown in (b). Use the intersect utility in the CALC menu to determine the point of intersection, as shown in (c).
The graph of y = 0.15x − 0.23 is above the graph of y = 8.2 − 0.6x for all values of x that lie to the right of 11.24. Therefore, the solution of 0.15x − 0.23 > 8.2 − 0.6x is ((11.24,infty) = {x : x > 11.24})
Exercise (PageIndex{24})
(−1.23x − 9.76 leq 1.44x + 22.8)
Exercise (PageIndex{25})
(0.5x^2 − 5 < 1.23 − 0.75x)
- Answer
-
To solve the inequality (0.5x^2 − 5 < 1.23 − 0.75x) graphically, start by loading the left- and right-hand sides of the inequality into Y1 and Y2, respectively, as shown in (a). Use the intersect utility in the CALC menu to determine the points of intersection, as shown in (b) and (c).
The graph of y = 0.5×2−5 is below the graph of y = 1.23−0.75x for all values of x that lie between −4.35867 and 2.8586701. Therefore, the solution of (0.5x^2−5 < 1.23−0.75x) is (−4.35867, 2.8586701) or {x : −4.35867 < x < 2.8586701}.
Exercise (PageIndex{26})
(4 − 0.5x^2 leq 0.72x − 1.34)
In Exercises (PageIndex{27})-(PageIndex{30}), perform each of the following tasks. Remember to use a ruler to draw all lines.
i. Make an accurate copy of the image on graph paper (label the graph with the letter f and label and scale each axis), drop a dashed vertical line through the x-intercept of the graph of f, then label and shade the solution of f(x) = 0 on the x-axis. Use set-builder notation to describe your solution.
ii. Make a second copy of the image on graph paper, drop a dashed, vertical line through the x-intercept of the graph of f, then label and shade the solution of f(x) > 0 on the x-axis. Use set-builder and interval notation to describe your solution set.
iii. Make a third copy of the image on graph paper, drop a dashed, vertical line through the x-intercept of the graph of f, then label and shade the solution of f(x) < 0 on the x-axis. Use set-builder and interval notation to describe your solution set.
Exercise (PageIndex{27})
- Answer
-
The graph of f intercepts the x-axis at x = −1. Therefore, the solution of f(x) = 0 is x = −1.
The graph of f lies above the x-axis for all values of x that lie to the right of −1. Therefore, the solution of f(x) > 0 is ((−1,infty) = {x : x > −1}).
The graph of f lies below the x-axis for all values of x that lie to the left of −1. Therefore, the solution of f(x) < 0 is ((−infty, −1) = {x : x < −1})
Exercise (PageIndex{28})
Exercise (PageIndex{29})
- Answer
-
The graph of f intercepts the x-axis at x = 2. Therefore, the solution of f(x) = 0 is x = 2.
The graph of f lies above the x-axis for all values of x that lie to the left of x = 2. Therefore, the solution of f(x) > 0 is ((−infty, 2) = {x : x < 2}).
The graph of f lies below the x-axis for all values of x that lie to the right of x = 2. Therefore, the solution of f(x) < 0 is ((2,infty) = {x : x > 2})
Exercise (PageIndex{30})
In Exercises (PageIndex{31})-(PageIndex{34}), perform each of the following tasks. Remember to use a ruler to draw all lines.
i. Make an accurate copy of the image on graph paper, drop dashed, vertical lines through the x-intercepts, then label and shade the solution of (f(x) geq 0) on the x-axis. Use set-builder and interval notation to describe your solution set.
ii. Make a second copy of the image on graph paper, drop dashed, vertical lines through the x-intercepts, then label and shade the solution of f(x) < 0 on the x-axis. Use set-builder and interval notation to describe your solution set.
Exercise (PageIndex{31})
- Answer
-
The graph of f intercepts the x-axis at x = −3 and x = 2. The graph of f lies above the x-axis for all values of x that lie between x = −3 and x = 2. Therefore, the solution of (f(x) geq 0) is ([−3, 2] = {x : −3 leq x leq 2}).
The graph of f lies below the x-axis for all values of x that lie to the left of x = −3 or to the right of x = 2. Therefore, the solution of f(x) < 0 is ((−infty, −3) cup (2,infty) = {x : x < −3 or x > 2}).
Exercise (PageIndex{32})
Exercise (PageIndex{33})
- Answer
-
The graph of f intercepts the x-axis at x = −2 and x = 1. The graph of f lies above the x-axis for all values of x that lie to the left x = −2 or to the right of x = 1. Therefore, the solution of (f(x) geq 0) is ((−infty, −2] cup [1,infty) = {x : x leq −2 or x geq 1}).
The graph of f lies below the x-axis for all values of x that lie between x = −2 and x = 1. Therefore, the solution of f(x) < 0 is ((−2, 1) = {x : −2 < x < 1}).
Exercise (PageIndex{34})
In Exercises (PageIndex{35})-(PageIndex{38}), perform each of the following tasks. Remember to use a ruler to draw all lines.
i. Load the given function f into the Y= menu of your calculator. Adjust the WINDOW parameters so that the x-intercept(s) of the graph of f is visible in the viewing window. Use the zero utility in the CALC menu of your calculator to determine the coordinates of the x-intercept(s) of the graph of f.
ii. Make an accurate copy of the image in your viewing window on your homework paper. Label and scale each axis with xmin, xmax, ymin, and ymax, and label the graph with its equation.
iii. Draw a dashed, vertical line through the x-intercept(s). Shade and label the solution of the inequality f(x) > 0 on the x-axis. Use both set-builder and interval notation to describe the solution set.
Exercise (PageIndex{35})
f(x) = −1.25x + 3.58
- Answer
-
To solve the inequality f(x) > 0 graphically, start by loading f(x) = −1.25x+3.58 into Y1. Use the zero utility in the CALC menu to determine the zero of f, as shown in (c).
The graph of f lies above the x-axis for all values of x that lie to the left of x = 2.864. Therefore, the solution of f(x) > 0 is ((−infty, 2.864) = {x : x < 2.864}). Answers may vary due to round-off error.
Exercise (PageIndex{36})
f(x) = 1.34x − 4.52
Exercise (PageIndex{37})
(f(x) = 1.25x^2 + 4x − 5.9125)
- Answer
-
To solve the inequality f(x) > 0 graphically, start by loading (f(x) = 1.25x^2 + 4x − 5.9125) into Y1. Use the zero utility in the CALC menu to determine the zeros of f, as shown in (b) and (c).
The graph of f lies above the x-axis for all values of x that lie to the left of x = −4.3 or to the right of x = 1.1. Therefore, the solution of f(x) > 0 is ((−infty, −4.3) cup (1.1,infty)) or ({x : x < −4.3 or x > 1.1}). Answers may vary due to round-off error.
Exercise (PageIndex{38})
(f(x) = −1.32x^2 − 3.96x + 5.9532)
In Exercises (PageIndex{39})-(PageIndex{42}), perform each of the following tasks. Remember to use a ruler to draw all lines.
i. Load the given function f into the Y= menu of your calculator. Adjust the WINDOW parameters so that the x-intercept(s) of the graph of f is visible in the viewing window. Use the zero utility in the CALC menu of your calculator to determine the coordinates of the x-intercept(s) of the graph of f.
ii. Make an accurate copy of the image in your viewing window on your homework paper. Label and scale each axis with xmin, xmax, ymin, and ymax, and label the graph with its equation.
iii. Draw a dashed, vertical line through the x-intercept(s). Shade and label the solution of the inequality (f(x) leq 0) on the x-axis. Use both set-builder and interval notation to describe the solution set.
Exercise (PageIndex{39})
f(x) = −1.45x − 5.6
- Answer
-
To solve the inequality (f(x) leq 0) graphically, start by loading f(x) = −1.45x−5.6 into Y1. Use the zero utility in the CALC menu to determine the zero of f, as shown in (c).
The graph of f intercepts the x-axis at x = −3.862069. The graph of f lies below the x-axis for all values of x that lie to the right of x = −3.862069. Therefore, the solution of (f(x) leq 0) is ([−3.862069,infty) = {x : x geq −3.862069}). Answers may vary due to roundoff error.
Exercise (PageIndex{40})
f(x) = 1.35x + 8.6
Exercise (PageIndex{41})
(f(x) = −1.11x^2 −5.9940x+1.2432)
- Answer
-
To solve the inequality (f(x) leq 0) graphically, start by loading (f(x) = −1.11x^2 − 5.9940x+1.2432) into Y1. Use the zero utility in the CALC menu to determine the zeros of f, as shown in (b) and (c).
The graph of f intercepts the x-axis at x = −5.6 and x = 0.2. The graph of f lies below the x-axis for all values of x that lie to the left of x = −5.6 or to the right of x = 0.2. Therefore, the solution of (f(x) leq 0) is ((−infty, −5.6] cup [0.2,infty)) or ({x : x leq −5.6 or x geq 0.2}). Answers may vary due to roundoff error.
Exercise (PageIndex{42})
(f(x) = 1.22x^2 − 6.3440x + 1.3176)
2.5 Exercises
Pictured below is the graph of a function f.
The table that follows evaluates the function f in the plot at key values of x. Notice the horizontal format, where the first point in the table is the ordered pair (−4, 0).
| x | -4 | -3 | 0 | 2 | 5 | 6 |
| f(x) | 0 | 4 | 4 | -4 | -4 | 0 |
Use the graph and the table to complete each of following tasks for Exercises (PageIndex{1}) – (PageIndex{10}).
i. Set up a coordinate system on graph paper. Label and scale each axis, then copy and label the original graph of f onto your coordinate system. Remember to draw all lines with a ruler.
ii. Use the original table to help complete the table for the given function in the exercise.
iii. Using a different colored pencil, plot the data from your completed table on the same coordinate system as the original graph of f. Use these points to help complete the graph of the given function in the exercise, then label this graph with its equation given in the exercise.
Exercise (PageIndex{1})
y= 2f(x).
| x | -4 | -3 | 0 | 2 | 5 | 6 |
| y |
- Answer
-
The original function table.
x -4 -3 0 2 5 6 f(x) 0 4 4 -4 -4 0
Evaluate the function y = 2f(x) at x = −4, −3, 0, 2, 5, and 6.
[y= 2f(−4) = 2(0) = 0 \ y = 2f(−3) = 2(4) = 8 \ y = 2f(0) = 2(4) = 8 \ y = 2f(2) = 2(−4) = −8 \ y = 2f(5) = 2(−4) = −8 \ y = 2f(6) = 2(0) = 0 ]
Points satisfying y = 2f(x).
| x | -4 | -3 | 0 | 2 | 5 | 6 |
| y | 0 | 8 | 8 | -8 | -8 | 0 |
Plot the points in the table to get the graph of y = 2f(x).
Note that multiplying by 2, as in y = 2f(x), stretches the graph of y = f(x) vertically by a factor of 2.
Exercise (PageIndex{2})
y = (1/2)f(x).
| x | -4 | -3 | 0 | 2 | 5 | 6 |
| y |
Exercise (PageIndex{3})
y = −f(x).
| x | -4 | -3 | 0 | 2 | 5 | 6 |
| y |
- Answer
-
The original function table.
x -4 -3 0 2 5 6 f(x) 0 4 4 -4 -4 0
Evaluate the function y = −f(x) at x = −4, −3, 0, 2, 5, and 6.
[y = −f(−4) = −(0) = 0 \ y = −f(−3) = −(4) = −4 \ y = −f(0) = −(4) = −4 \ y = −f(2) = −(−4) = 4 \ y = −f(5) = −(−4) = 4 \ y = −f(6) = −(0) = 0]
Points satisfying y = −f(x).
| x | -4 | -3 | 0 | 2 | 5 | 6 |
| y | 0 | -4 | -4 | 4 | 4 | 0 |
Plot the points in the table to get the graph of y = −f(x).
Note that negating the function, as in y = −f(x), reflects the graph of y = f(x) across the x-axis.
Exercise (PageIndex{4})
y = f(x) − 2.
| x | -4 | -3 | 0 | 2 | 5 | 6 |
| y |
Exercise (PageIndex{5})
y = f(x) + 4.
| x | -4 | -3 | 0 | 2 | 5 | 6 |
| y |
- Answer
-
The original function table.
x -4 -3 0 2 5 6 f(x) 0 4 4 -4 -4 0
Evaluate the function y = f(x) + 4 at x = −4, −3, 0, 2, 5, and 6.
[y = f(−4) + 4 = (0) + 4 = 4 \y = f(−3) + 4 = (4) + 4 = 8 \y = f(0) + 4 = (4) + 4 = 8 \y = f(2) + 4 = (−4) + 4 = 0 \y = f(5) + 4 = (−4) + 4 = 0 \y = f(6) + 4 = (0) + 4 = 4]
Points satisfying y = f(x) + 4.
| x | -4 | -3 | 0 | 2 | 5 | 6 |
| y | 4 | 8 | 8 | 0 | 0 | 4 |
Plot the points in the table to get the graph of y = f(x) + 4.
Note that adding 4, as in y = f(x) + 4, translates the graph of y = f(x) upwards 4 units.
Exercise (PageIndex{6})
y = −2f(x)
| x | -4 | -3 | 0 | 2 | 5 | 6 |
| y |
Exercise (PageIndex{7})
y = (−1/2)f(x)
| x | -4 | -3 | 0 | 2 | 5 | 6 |
| y |
- Answer
-
The original function table.
x -4 -3 0 2 5 6 f(x) 0 4 4 -4 -4 0
Evaluate the function y = (−1/2)f(x) at x = −4, −3, 0, 2, 5, and 6.
[y = (−1/2)f(−4) = (−1/2)(0) = 0 \ y = (−1/2)f(−3) = (−1/2)(4) = −2 \ y = (−1/2)f(0) = (−1/2)(4) = −2 \ y = (−1/2)f(2) = (−1/2)(−4) = 2 \ y = (−1/2)f(5) = (−1/2)(−4) = 2 \ y = (−1/2)f(6) = (−1/2)(0) = 0 ]
Points satisfying y = (−1/2)f(x).
| x | -4 | -3 | 0 | 2 | 5 | 6 |
| y | 0 | -2 | -2 | 2 | 2 | 0 |
Plot the points in the table to get the graph of y = (−1/2)f(x).
Note that multiplying by −1/2, as in y = (−1/2)f(x), compresses the graph of y = f(x) vertically by a factor of 2, then reflects the result across the x-axis.
Exercise (PageIndex{8})
y = −f(x) + 3.
| x | -4 | -3 | 0 | 2 | 5 | 6 |
| y |
Exercise (PageIndex{9})
y = −f(x) − 2
| x | -4 | -3 | 0 | 2 | 5 | 6 |
| y |
- Answer
-
The original function table.
x -4 -3 0 2 5 6 f(x) 0 4 4 -4 -4 0
Evaluate the function y = −f(x) − 2 at x = −4, −3, 0, 2, 5, and 6.
[y = −f(−4) − 2 = −(0) − 2 = −2 \ y = −f(−3) − 2 = −(4) − 2 = −6\ y = −f(0) − 2 = −(4) − 2 = −6 \y = −f(2) − 2 = −(−4) − 2 = 2 \y = −f(5) − 2 = −(−4) − 2 = 2 \y = −f(6) − 2 = −(0) − 2 = −2 ]
Points satisfying y = −f(x) − 2.
| x | -4 | -3 | 0 | 2 | 5 | 6 |
| y | -2 | -6 | -6 | 2 | 2 | -2 |
Plot the points in the table to get the graph of y = −f(x) − 2.
Note that negating then subtracting 2, as in y = −f(x) − 2, first reflects the graph of y = f(x) across the x-axis, then translates the resulting reflection 2 units downward.
Exercise (PageIndex{10})
y = (−1/2)f(x) + 3
| x | -4 | -3 | 0 | 2 | 5 | 6 |
| y |
Exercise (PageIndex{11})
Use your graphing calculator to draw the graph of (y = sqrt{x}). Then, draw the graph of (y = sqrt{x}). In your own words, explain what you learned from this exercise.
- Answer
-
First, draw the graph of (y = sqrt{x}).
The graph of (y = -sqrt{x}) is a reflection of the graph of (y = sqrt{x}) across the x-axis.
Negating a function appears to reflect the graph of the function across the x-axis.
Exercise (PageIndex{12})
Use your graphing calculator to draw the graph of y = |x|. Then, draw the graph of y = −|x|. In your own words, explain what you learned from this exercise.
Exercise (PageIndex{13})
Use your graphing calculator to draw the graph of (y = x^2). Then, in succession, draw the graphs of (y = x^2−2), (y = x^2−4), and (y = x^2 − 6). In your own words, explain what you learned from this exercise.
- Answer
-
First, draw the graph of (y = x^2).
Subtracting 2 (as in (y = x^2-2)) translates the graph of (y = x^2) two units downward in the y-direction.
Similarly, subtracting 4 and 6 translates the graph of (y = x^2) four units and 6 units downward, respectively
In general, if c is positive, then the graph of y = f(x) − c is obtained by translating the graph of y = f(x) downward c units.
Exercise (PageIndex{14})
Use your graphing calculator to draw the graph of (y = x^2). Then, in succession, draw the graphs of (y = x^2+2), (y = x^2+4), and (y = x^2 + 6). In your own words, explain what you learned from this exercise.
Exercise (PageIndex{15})
Use your graphing calculator to draw the graph of y = |x|. Then, in succession, draw the graphs of y = 2|x|, y = 3|x|, and y = 4|x|. In your own words, explain what you learned from this exercise.
- Answer
-
First, draw the graph of y = |x|.
Multiplying by 2, as in y = 2|x|, stretches the graph of y = |x| vertically by a factor of 2.
Similarly, multiplying by 3 and 4, as in y = 3|x| and y = 4|x|, stretches the graph of y = |x| vertically by factors of 3 and 4, respectively.
In general, if a > 1, then the graph of y = af(x) is obtained by stretching the graph of y = f(x) vertically by a factor of a.
Exercise (PageIndex{16})
Use your graphing calculator to draw the graph of y = |x|. Then, in succession, draw the graphs of y = (1/2)|x|, y = (1/3)|x|, and y = (1/4)|x|. In your own words, explain what you learned from this exercise.
Pictured below is the graph of a function f. In Exercises (PageIndex{17})-(PageIndex{22}), use this graph to perform each of the following tasks.
i. Set up a coordinate system on a sheet of graph paper. Label and scale each axis. Make an exact copy of the graph of f on your coordinate system. Remember to draw all lines with a ruler.
ii. In the narrative, a shadow box at the end of the section summarizes the concepts and technique of vertical scaling, vertical reflection, and vertical translation. Use the shortcut ideas presented in this summary shadow box to draw the graphs of the functions that follow without using tables.
iii. Use a different colored pencil to draw the graph of the function given in the exercise. Label this graph with its equation. Be sure that key points are accurately plotted. In each exercise, please plot exactly two plots per coordinate system, the graph of original function f and the graph of the function in the exercise.
Exercise (PageIndex{17})
y = (1/2)f(x).
- Answer
-
To obtain the plot of y = (1/2)f(x), simply multiply the y-value of each point on the graph of y = f(x) by 1/2, keeping the x-value the same.
Note that multiplying by 1/2, as in y = (1/2)f(x), compresses the graph of y = f(x) vertically by a factor of 2.
Exercise (PageIndex{18})
y = 2f(x).
Exercise (PageIndex{19})
y= −f(x).
- Answer
-
To obtain the plot of y = −f(x), simply negate the y-value of each point on the graph of y = f(x).
Note that negating, as in y = −f(x), reflects the graph of y = f(x) across the x-axis.
Exercise (PageIndex{20})
y = f(x) − 1
Exercise (PageIndex{21})
y = f(x) + 3.
- Answer
-
To obtain the plot of y = f(x) + 3, simply add 3 to the y-value of each point on the graph of y = f(x).
Note that adding 3, as in y = f(x) + 3, translates the graph of y = f(x) upwards 3 units.
Exercise (PageIndex{22})
y = f(x) − 4
Pictured below is the graph of a function f. In Exercises (PageIndex{23})-(PageIndex{28}), use this graph to perform each of the following tasks.
i. Set up a coordinate system on a sheet of graph paper. Label and scale each axis. Make an exact copy of the graph of f on your coordinate system. Remember to draw all lines with a ruler.
ii. In the narrative, a shadow box at the end of the section summarizes the concepts and technique of vertical scaling, vertical reflection, and vertical translation. Use the shortcut ideas presented in this summary shadow box to draw the graphs of the functions that follow without using tables.
iii. Use a different colored pencil to draw the graph of the function given in the exercise. Label this graph with its equation. Be sure that key points are accurately plotted. In each exercise, please plot exactly two plots per coordinate system, the graph of original function f and the graph of the function in the exercise.
Exercise (PageIndex{23})
y = 2f(x)
- Answer
-
To obtain the plot of y = 2f(x), simply multiply the y-value of each point of y = f(x) by 2.
Note that multiplying by 2, as in y = 2f(x), stretches the graph of y = f(x) vertically by a factor of 2.
Exercise (PageIndex{24})
y = (1/2)f(x)
Exercise (PageIndex{25})
y = −f(x).
- Answer
-
To obtain the plot of y = −f(x), simply negate the y-value of each point on the graph of y = f(x).
Note that negating a function, as in y = −f(x), reflects the graph of y = f(x) across the x-axis.
Exercise (PageIndex{26})
y = f(x) + 3
Exercise (PageIndex{27})
y = f(x) − 2
- Answer
-
To obtain the plot of y = f(x) − 2, simply subtract 2 from the y-value of each point on the graph of y = f(x).
Note that subtracting 2, as in y = f(x)−2, translates the graph of y = f(x) downwards 2 units.
Exercise (PageIndex{28})
y = f(x) − 1
Pictured below is the graph of a function f. In Exercises (PageIndex{29})-(PageIndex{34}), use this graph to perform each of the following tasks.
i. Set up a coordinate system on a sheet of graph paper. Label and scale each axis. Make an exact copy of the graph of f on your coordinate system. Remember to draw all lines with a ruler.
ii. In the narrative, a shadow box at the end of the section summarizes the concepts and technique of vertical scaling, vertical reflection, and vertical translation. Use the shortcut ideas presented in this summary shadow box to draw the graphs of the functions that follow without using tables.
iii. Use a different colored pencil to draw the graph of the function given in the exercise. Label this graph with its equation. Be sure that key points are accurately plotted. In each exercise, please plot exactly two plots per coordinate system, the graph of original function f and the graph of the function in the exercise.
Exercise (PageIndex{29})
y = (−1/2)f(x)
- Answer
-
We proceed in two steps:
1. First, multiply the y-value of each point on the graph of y = f(x) by 1/2 to produce the graph of y = (1/2)f(x) in (b). This compresses the graph of y = f(x) by a factor of 2.
2. Secondly, multiply the y-value of each point on the graph of y = (1/2)f(x) by −1 to produce the graph of y = (−1/2)f(x) in (c). This reflects the graph of y = (1/2)f(x) across the x-axis.
Exercise (PageIndex{30})
y = −2f(x).
Exercise (PageIndex{31})
y = −f(x) + 2
- Answer
-
We proceed in two steps:
1. First, multiply the y-value of each point on the graph of y = f(x) by −1 to produce the graph of y = −f(x) in (b). This reflects the graph of y = f(x) across the x-axis.
2. Secondly, add 2 to the y-value of each point on the graph of y = −f(x) to produce the graph of y = −f(x) + 2 in (c). This shifts the graph of y = −f(x) upward 2 units.
Exercise (PageIndex{32})
y = −f(x) − 3
Exercise (PageIndex{33})
y = 2f(x) − 3.
- Answer
-
We proceed in two steps:
1. First, multiply the y-value of each point on the graph of y = f(x) by 2 to produce the graph of y = 2f(x) in (b). This stretches the graph of y = f(x) vertically by a factor of 2.
2. Secondly, subtract 3 from the y-value of each point on the graph of y = 2f(x) to produce the graph of y = 2f(x) − 3 in (c). This shifts the graph of y = 2f(x) downward 3 units.
Exercise (PageIndex{34})
y = (−1/2)f(x) + 1
2.6 Exercises
Pictured below is the graph of a function f.
The table that follows evaluates the function f in the plot at key values of x. Notice the horizontal format, where the first point in the table is the ordered pair (−6, 0).
| x | -6 | -4 | -2 | 0 | 2 | 4 |
| f(x) | 0 | 4 | 4 | 0 | -2 | 0 |
Use the graph and the table to complete each of following tasks for Exercises (PageIndex{1}) – (PageIndex{10}).
i. Set up a coordinate system on graph paper. Label and scale each axis, then copy and label the original graph of f onto your coordinate system. Remember to draw all lines with a ruler.
ii. Use the original table to help complete the table for the given function in the exercise.
iii. Using a different colored pencil, plot the data from your completed table on the same coordinate system as the original graph of f. Use these points to help complete the graph of the given function in the exercise, then label this graph with its equation given in the exercise.
Exercise (PageIndex{1})
y = f(2x).
| x | -3 | -2 | -1 | 0 | 1 | 2 |
| y |
- Answer
-
The original function table.
x -6 -4 -2 0 2 4 f(x) 0 4 4 0 -2 0
Evaluate the function y = f(2x) at x = −3, −2, −1, 0, 1, and 2.
[y = f(2(−3)) = f(−6) = 0 \ y = f(2(−2)) = f(−4) = 4 \ y = f(2(−1)) = f(−2) = 4 \ y = f(2(0)) = f(0) = 0 \ y = f(2(1)) = f(2) = −2 \ y = f(2(2)) = f(4) = 0 ]
Points satisfying y = f(2x).
| x | -3 | -2 | -1 | 0 | 1 | 2 |
| y | 0 | 4 | 4 | 0 | -2 | 0 |
Plot the points in the table to get the graph of y = f(2x).
Note that replacing x with 2x, as in y = f(2x), compresses the graph of y = f(x) horizontally by a factor of 2.
Exercise (PageIndex{2})
y = f((1/2)x).
| x | -12 | -8 | -4 | 0 | 4 | 8 |
| y |
Exercise (PageIndex{3})
y = f(−x).
| x | -4 | -2 | 0 | 2 | 4 | 6 |
| y |
- Answer
- The original function table.
-
x -6 -4 -2 0 2 4 f(x) 0 4 4 0 -2 0
Evaluate the function y = f(−x) at x = −4, −2, 0, 2, 4, and 6.
[y = f(−(−4)) = f(4) = 0 \ y = f(−(−2)) = f(2) = −2 \ y = f(−(0)) = f(0) = 0 \ y = f(−(2)) = f(−2) = 4 \ y = f(−(4)) = f(−4) = 4 \ y = f(−(6)) = f(−6) = 0 ]
Points satisfying y = f(−x).
| x | -4 | -2 | 0 | 2 | 4 | 6 |
| y | 0 | -2 | 0 | 4 | 4 | 0 |
Plot the points in the table to get the graph of y = f(−x).
Note that replacing x with −x, as in y = f(−x), reflects the graph of y = f(x) across the y-axis.
Exercise (PageIndex{4})
y = f(x + 3).
| x | -9 | -7 | -5 | -3 | -1 | 1 |
| y |
Exercise (PageIndex{5})
y = f(x − 1).
| x | -5 | -3 | -1 | 1 | 3 | 5 |
| y |
- Answer
-
The original function table.
x -6 -4 -2 0 2 4 f(x) 0 4 4 0 -2 0
Evaluate the function y = f(x − 1) at x = −5, −3, −1, 1, 3, and 5.
[y = f((−5) − 1) = f(−6) = 0 \ y = f((−3) − 1) = f(−4) = 4 \ y = f((−1) − 1) = f(−2) = 4 \ y = f((1) − 1) = f(0) = 0 \ y = f((3) − 1) = f(2) = −2 \ y = f((5) − 1) = f(4) = 0]
Points satisfying y = f(x − 1).
| x | -5 | -3 | -1 | 1 | 3 | 5 |
| y | 0 | 4 | 4 | 0 | -2 | 0 |
Plot the points in the table to get the graph of y = f(x − 1).
Note that replacing x with x − 1, as in y = f(x − 1), translates the graph of y = f(x) horizontally 1 unit to the right.
Exercise (PageIndex{6})
y = f(−2x).
| x | -2 | -1 | 0 | 1 | 2 | 3 |
| y |
Exercise (PageIndex{7})
y = f((−1/2)x).
| x | -8 | -4 | 0 | 4 | 8 | 12 |
| y |
- Answer
-
The original function table.
x -6 -4 -2 0 2 4 f(x) 0 4 4 0 -2 0
Evaluate the function y = f((−1/2)x) at x = −8, −4, 0, 4, 8, and 12.
[y = f((−1/2)(−8)) = f(4) = 0 \ y = f((−1/2)(−4)) = f(2) = −2 \ y = f((−1/2)(0)) = f(0) = 0 \ y = f((−1/2)(4)) = f(−2) = 4 \ y = f((−1/2)(8)) = f(−4) = 4 \ y = f((−1/2)(12)) = f(−6) = 0]
Points satisfying y = f((−1/2)x).
| x | -8 | -4 | 0 | 4 | 8 | 12 |
| y | 0 | -2 | 0 | 4 | 4 | 0 |
Plot the points in the table to get the graph of y = f((−1/2)x).
Note that replacing x with (−1/2)x, as in y = f((−1/2)x), stretches the graph by a factor of 2, then reflects the result across the y-axis.
Exercise (PageIndex{8})
y= f(−x − 2)
| x | -6 | -4 | -2 | 0 | 2 | 4 |
| y |
Exercise (PageIndex{9})
y= f(−x + 1).
| x | -3 | -1 | 1 | 3 | 5 | 7 |
| y |
- Answer
-
The original function table.
x -6 -4 -2 0 2 4 f(x) 0 4 4 0 -2 0
Evaluate the function y = f(−x + 1) at x = −3, −1, 1, 3, 5, and 7.
[y = f(−(−3) + 1) = f(4) = 0 \ y = f(−(−1) + 1) = f(2) = −2 \ y = f(−(1) + 1) = f(0) = 0 \ y = f(−(3) + 1) = f(−2) = 4 \ y = f(−(5) + 1) = f(−4) = 4 \ y = f(−(7) + 1) = f(−6) = 0 ]
Points satisfying y = f(−x + 1).
| x | -3 | -1 | 1 | 3 | 5 | 7 |
| y | 0 | -2 | 0 | 4 | 4 | 0 |
Plot the points in the table to get the graph of y = f(−x + 1).
Note that y = f(−x + 1) is the same as y = f(−(x − 1)). If we replace x with −x to get y = f(−x), then x in this last result with x − 1 to get y = f(−(x − 1)), this has the effect of first reflecting the graph of y = f(x) across the y-axis, then shifting the result to the right 1 unit.
Exercise (PageIndex{10})
y = f(−x/4).
| x | -16 | -8 | 0 | 8 | 16 | 24 |
| y |
Exercise (PageIndex{11})
Use your graphing calculator to draw the graph of (y = sqrt{x}). Then, draw the graph of (y = sqrt{-x}). In your own words, explain what you learned from this exercise.
- Answer
-
First, draw the graph of (y = sqrt{x}).
The graph of (y = sqrt{-x}) is a reflection of the graph of (y = sqrt{x}) across the y-axis.
Replacing x with −x, as in y = f(−x), reflects the graph of y = f(x) across the y-axis.
Exercise (PageIndex{12})
Use your graphing calculator to draw the graph of y = |x|. Then, draw the graph of y = | − x|. In your own words, explain what you learned from this exercise.
Exercise (PageIndex{13})
Use your graphing calculator to draw the graph of (y = x^2). Then, in succession, draw the graphs of (y = (x − 2)^2, y = (x − 4)^2), and (y = (x − 6)^2). In your own words, explain what you learned from this exercise.
- Answer
-
First, draw the graph of (y = x^2).
Replacing x with x − 2 translates the graph of (y = x^2) two units to the right in the horizontal direction.
Similarly, replacing x with x − 4 and x − 6 translates the graph of (y = x^2) four units and 6 units to the right, respectively.
In general, if c is positive, then the graph of y = f(x − c) is obtained by translating the graph of y = f(x) to the right c units.
Exercise (PageIndex{14})
Use your graphing calculator to draw the graph of (y = x^2). Then, in succession, draw the graphs of (y = (x + 2)^2, y = (x + 4)^2), and (y = (x + 6)^2). In your own words, explain what you learned from this exercise.
Exercise (PageIndex{15})
Use your graphing calculator to draw the graph of y = |x|. Then, in succession, draw the graphs of y = |2x|, y = |3x|, and y = |4x|. In your own words, explain what you learned from this exercise.
- Answer
-
First, draw the graph of y = |x|.
Replacing x with 2x compresses the graph of y = |x| by a factor of 2 in the horizontal direction.
Similarly, replacing x with 3x and 4x by a factor of 3 and 4 in the horizontal direction, respectively.
In general, if a > 1, then the graph of y = f(ax) is obtained by compressing the graph of y = f(x) by a factor of a in the horizontal direction.
Exercise (PageIndex{16})
Use your graphing calculator to draw the graph of y = |x|. Then, in succession, draw the graphs of y = |(1/2)x|, y = |(1/3)x|, and y = |(1/4)x|. In your own words, explain what you learned from this exercise.
Pictured below is the graph of a function f. In Exercises (PageIndex{17})-(PageIndex{22}), use this graph to perform each of the following tasks.
i. Set up a coordinate system on a sheet of graph paper. Label and scale each axis. Make an exact copy of the graph of f on your coordinate system. Remember to draw all lines with a ruler.
ii. In the narrative, a shadow box at the end of the section summarizes the concepts and technique of horizontal scaling, horizontal reflection, and horizontal translation. Use the shortcut ideas presented in this summary shadow box to draw the graphs of the functions that follow without using tables.
iii. Use a different colored pencil to draw the graph of the function given in the exercise. Label this graph with its equation. Be sure that key points are accurately plotted. In each exercise, please plot exactly two plots per coordinate system, the graph of original function f and the graph of the function in the exercise.
Exercise (PageIndex{17})
y = f(2x).
- Answer
-
To obtain a plot for y = f(2x), take each point on the graph of y = f(x) and divide its x-value by 2, keeping the y-value the same.
Note that replacing x with 2x, as in y = f(2x), compresses the graph of y = f(x) in the horizontal direction by a factor of 2.
Exercise (PageIndex{18})
y = f((1/2)x).
Exercise (PageIndex{19})
y = f(−x).
- Answer
-
To obtain a plot of y = f(−x), take each point on the graph of y = f(x) and negate its x-value, keeping the y-value the same.
Note that replacing x with −x, as in y = f(−x), reflects the graph of f across the y-axis.
Exercise (PageIndex{20})
y = f(x − 1).
Exercise (PageIndex{21})
y = f(x + 3).
- Answer
-
To obtain a plot of y = f(x + 3), take each point on the graph of y = f(x) and subtract 3 from its x-value, keeping the y-value the same.
Note that replacing x with x + 3, as in y = f(x + 3), translates the graph of y = f(x) to the left 3 units.
Exercise (PageIndex{22})
y = f(x − 2).
Pictured below is the graph of a function f. In Exercises (PageIndex{23})-(PageIndex{28}), use this graph to perform each of the following tasks.
i. Set up a coordinate system on a sheet of graph paper. Label and scale each axis. Make an exact copy of the graph of f on your coordinate system. Remember to draw all lines with a ruler.
ii. In the narrative, a shadow box at the end of the section summarizes the concepts and technique of horizontal scaling, horizontal reflection, and horizontal translation. Use the shortcut ideas presented in this summary shadow box to draw the graphs of the functions that follow without using tables.
iii. Use a different colored pencil to draw the graph of the function given in the exercise. Label this graph with its equation. Be sure that key points are accurately plotted. In each exercise, please plot exactly two plots per coordinate system, the graph of original function f and the graph of the function in the exercise.
Exercise (PageIndex{23})
y = f(2x).
- Answer
-
To obtain a plot of y = f(2x), take each point on the graph of y = f(x) and divide its x-value by 2, keeping the y-value the same.
Replacing x with 2x, as in y = f(2x), compresses the graph of y = f(x) horizontally by a factor of 2.
Exercise (PageIndex{24})
y = f((1/2)x).
Exercise (PageIndex{25})
y = f(−x).
- Answer
-
To obtain a plot of y = f(−x), take each point on the graph of y = f(x) and negate its x-value, keeping the same y-value.
Replacing x with −x, as in y = f(−x), reflects the graph of y = f(x) across the y-axis.
Exercise (PageIndex{26})
y = f(x + 3)
Exercise (PageIndex{27})
y= f(x − 2).
- Answer
-
To obtain a plot of y = f(x − 2), take each point on the graph of y = f(x) and add 2 to its x-value, keeping its y-value the same.
Replacing x with x − 2, as in y = f(x − 2), shifts the graph of y = f(x) to the right 2 units.
Exercise (PageIndex{28})
y = f(x + 1).