Question

which function in vertex form is equivalent to $f(x)=x^{2}+8 - 16x$?
$\bigcirc$ $f(x)=(x - 8)^{2}-56$
$\bigcirc$ $f(x)=(x - 4)^{2}+0$
$\bigcirc$ $f(x)=(x + 8)^{2}-72$
$\bigcirc$ $f(x)=(x + 4)^{2}-32$

Explanation:

Step1: Rearrange the quadratic terms

$f(x) = x^2 - 16x + 8$

Step2: Complete the square for $x^2-16x$

Take half of $-16$: $\frac{-16}{2}=-8$, square it: $(-8)^2=64$

Step3: Rewrite the function

$f(x) = (x^2 - 16x + 64) + 8 - 64$

Step4: Simplify to vertex form

$f(x) = (x - 8)^2 - 56$

Answer:

$f(x) = (x - 8)^2 - 56$