The correct answer is [tex]\mathrm{Parabola\:vertex\:given}\:y=8x^2-64x:\quad \mathrm{Minimum}\space\left(4,\:-128\right)[/tex]
Steps:
[tex]\mathrm{For\:a\:parabola}\:ax^2+bx+c\:\mathrm{the\:vertex's\:x\:equals}\:\frac{-b}{2a}
\:a=8,\:b=-64
x=\frac{-\left(-64\right)}{2\cdot \:8}[/tex]
[tex]\mathrm{Simplify}
x=4[/tex]
[tex]\mathrm{Plug\:in}\:x=4\:\mathrm{to\:find\:the\:y\:value}
y=8\cdot \:4^2-64\cdot \:4[/tex]
[tex]\mathrm{Simplify}
y=-128[/tex]
[tex]\mathrm{Therefore\:the\:parabola\:vertex\:is}
\left(4,\:-128\right)[/tex]
[tex]\mathrm{Minimum}\space\left(4,\:-128\right)[/tex]
