Question 1

What is the vertex of g(x)=8x^2-64x

Question 2

The correct answer is
$\mathrm{Parabola\:vertex\:given}\:y=8x^2-64x:\quad \mathrm{Minimum}\space\left(4,\:-128\right)$

Steps:

$\mathrm{For\:a\:parabola}\:ax^2+bx+c\:\mathrm{the\:vertex's\:x\:equals}\:(-b)/(2a) \:a=8,\:b=-64x=(-\left(-64\right))/(2\cdot \:8)$

$\mathrm{Simplify}x=4$

$\mathrm{Plug\:in}\:x=4\:\mathrm{to\:find\:the\:y\:value}y=8\cdot \:4^2-64\cdot \:4$

$\mathrm{Simplify}y=-128$

$\mathrm{Therefore\:the\:parabola\:vertex\:is}\left(4,\:-128\right)$

$\mathrm{Minimum}\space\left(4,\:-128\right)$

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