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f(x)=4x^(2)+64 x+262
The function
g is defined by
g(x)=f(x+5). For what value of
x does
g(x) reach its minimum?
A. -13
B. -8
C. -5
D. -3

$f(x)=4 x^{2}+64 x+262$ $\newline$ The function $g$ is defined by $g(x)=f(x+5)$ . For what value of $x$ does $g(x)$ reach its minimum? $\newline$ A. $-13$ $\newline$ B. $-8$ $\newline$ C. $-5$ $\newline$ D. $-3$

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Transformations of quadratic functions

Full solution

Q.

f(x)=4 x^{2}+64 x+262

\newline

The function

g

is defined by

g(x)=f(x+5)

. For what value of

x

does

g(x)

reach its minimum?

\newline

A.

-13

\newline

B.

-8

\newline

C.

-5

\newline

D.

-3

Substitute and Define $g(x)$ : $g(x)$ is defined as $f(x+5)$ , so substitute $x+5$ into $f(x)$ . $\newline$ $g(x) = f(x+5) = 4(x+5)^2 + 64(x+5) + 262$
Expand and Simplify: Expand the square and simplify. $\newline$ $g(x) = 4(x^2 + 10x + 25) + 64x + 320 + 262$
Distribute and Combine Terms: Distribute the $4$ and combine like terms. $g(x) = 4x^2 + 40x + 100 + 64x + 320 + 262$ $g(x) = 4x^2 + 104x + 682$
Find Vertex Form: The vertex form of a quadratic function is $a(x-h)^2 + k$ , where $(h, k)$ is the vertex. $\newline$ The x-coordinate of the vertex, $h$ , is given by $-b/(2a)$ . $\newline$ For $g(x) = 4x^2 + 104x + 682$ , $a = 4$ and $b = 104$ . $\newline$ $h = -104/(2\cdot4)$ $\newline$ $h = -104/8$ $\newline$ $h = -13$
Calculate Vertex Coordinates: The minimum value of $g(x)$ occurs at $x = h$ . So, the minimum value of $g(x)$ occurs at $x = -13$ .

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