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

Explanation:

Step1: Identify coefficients

For the quadratic function $g(x)=8x^{2}-64x$, where $a = 8$, $b=-64$, $c = 0$.

Step2: Find x - coordinate of vertex

Use the formula $x=-\frac{b}{2a}$. Substitute $a = 8$ and $b=-64$ into it: $x=-\frac{-64}{2\times8}=\frac{64}{16}=4$.

Step3: Find y - coordinate of vertex

Substitute $x = 4$ into $g(x)=8x^{2}-64x$. Then $g(4)=8\times4^{2}-64\times4=8\times16 - 256=128 - 256=-128$.

Answer:

$(4,-128)$