Question

what is the vertex of $g(x) = 8x^2 - 64x$?
○ $(4, -128)$
○ $(-4, -128)$
○ $(4, -16)$
○ $(-4, -16)$

Explanation:

Step1: Find x-coordinate of vertex

For $g(x)=ax^2+bx$, $x=-\frac{b}{2a}$. Here $a=8$, $b=-64$, so $x=-\frac{-64}{2\times8}=4$.

Step2: Find y-coordinate of vertex

Substitute $x=4$ into $g(x)$: $g(4)=8(4)^2-64(4)=8\times16-256=128-256=-128$.

Answer:

(4, -128)