which function in vertex form is equivalent to $f(x)=x^{2}+8 - 16x$?\n$f(x)=(x - 8)^{2}-56$\n$f(x)=(x…

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

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

Answer

Explanation:

Step1: Rewrite the given function

Given $f(x)=x^{2}+8 - 16x$, rewrite it as $f(x)=x^{2}-16x + 8$.

Step2: Complete the square

For the quadratic expression $x^{2}-16x$, take half of the coefficient of $x$ ($\frac{-16}{2}=-8$), then square it ($(-8)^{2}=64$). Add and subtract 64 in the function: $f(x)=(x^{2}-16x + 64)-64 + 8$.

Step3: Factor the perfect - square trinomial

$x^{2}-16x + 64=(x - 8)^{2}$, so $f(x)=(x - 8)^{2}-56$.

Answer:

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