Question

$f(x)=x^{2}+8x+12$
$-7,-4$
$boc = \underline{\quad\quad}$

Explanation:

Step1: Rewrite in vertex form

Complete the square for $f(x)=x^2+8x+12$.
$f(x)=(x^2+8x+16)-16+12=(x+4)^2-4$

Step2: Identify critical points

The vertex of the parabola is at $x=-4$, which is the right endpoint of the interval $[-7,-4]$. Evaluate $f(x)$ at the endpoints:
At $x=-4$: $f(-4)=(-4+4)^2-4=-4$
At $x=-7$: $f(-7)=(-7)^2+8(-7)+12=49-56+12=5$

Step3: Find range of the function

On the interval $[-7,-4]$, the parabola opens upward, so it decreases from $x=-7$ to $x=-4$. The minimum value is $-4$ and the maximum value is $5$.

Answer:

$[-4, 5]$