Question

22 mark for review
$f(x) = 4x^2 + 64x + 202$
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 $-3$

Explanation:

Step1: Find vertex of \( f(x) \)

For a quadratic \( ax^2 + bx + c \), vertex \( x \)-coordinate is \( -\frac{b}{2a} \). For \( f(x) = 4x^2 + 64x + 202 \), \( a = 4 \), \( b = 64 \). So \( x = -\frac{64}{2\times4} = -8 \). Thus, \( f(x) \) has minimum at \( x = -8 \).

Step2: Analyze \( g(x) = f(x + 5) \)

The function \( g(x) \) is a horizontal shift of \( f(x) \): replacing \( x \) with \( x + 5 \) shifts \( f(x) \) left by 5 units. To find the minimum of \( g(x) \), we solve \( x + 5 = -8 \) (since the minimum of \( f \) is at \( x = -8 \), so \( g(x) \) reaches minimum when \( x + 5 = -8 \)).

Step3: Solve for \( x \)

\( x + 5 = -8 \implies x = -8 - 5 = -13 \)? Wait, no—wait, wait. Wait, no: Wait, \( g(x) = f(x + 5) \). So the input to \( f \) is \( x + 5 \). The minimum of \( f \) is at \( x = -8 \) (input to \( f \) is \( -8 \)). So set \( x + 5 = -8 \implies x = -8 - 5 = -13 \)? Wait, but let's check again. Wait, no—wait, maybe I mixed up. Wait, \( f(x) \) has minimum at \( x = -8 \). Then \( g(x) = f(x + 5) \): so when does \( x + 5 = -8 \)? \( x = -13 \)? Wait, but let's verify with vertex form.

Alternative: Complete the square for \( f(x) \). \( f(x) = 4(x^2 + 16x) + 202 = 4(x^2 + 16x + 64 - 64) + 202 = 4((x + 8)^2 - 64) + 202 = 4(x + 8)^2 - 256 + 202 = 4(x + 8)^2 - 54 \). So \( f(x) = 4(x + 8)^2 - 54 \), minimum at \( x = -8 \). Then \( g(x) = f(x + 5) = 4((x + 5) + 8)^2 - 54 = 4(x + 13)^2 - 54 \). So \( g(x) \) is a quadratic with vertex at \( x = -13 \), so minimum at \( x = -13 \). Wait, but earlier step: when \( x + 5 = -8 \), \( x = -13 \). Yes, that's correct. So the minimum of \( g(x) \) is at \( x = -13 \).

Answer:

A. -13