Question

What is the zero of y=-4x^2+8x-28

Answer 1

$Solution, y-4x^2+8x-28=0:\quad \mathrm{Parabola\:with\:vertex\:at}\:\left(h,\:k\right)=\left(1,\:24\right),\:\mathrm{and\:focal\:length}\:|p|=\frac{1}{16}$

$Steps:$

$\mathrm{Rewrite}\:y-4x^2+8x-28=0\:\mathrm{in\:the\:standard\:form}$

$\mathrm{Rewrite\:as}, y=4x^2-8x+28$

$\mathrm{Complete\:the\:square}\:4x^2-8x+28:\quad 4\left(x-1\right)^2+24, y=4\left(x-1\right)^2+24$

$\mathrm{Subtract\:}24\mathrm{\:from\:both\:sides}, y-24=4\left(x-1\right)^2$

$\mathrm{Divide\:by\:coefficient\:of\:square\:terms:\:}4, \frac{1}{4}\left(y-24\right)=\left(x-1\right)^2$

$\mathrm{Rewrite\:in\:standard\:form}, 4\cdot \frac{1}{16}\left(y-24\right)=\left(x-1\right)^2$

$\mathrm{Therefore\:parabola\:properties\:are:} \left(h,\:k\right)=\left(1,\:24\right),\:p=\frac{1}{16}$

$\mathrm{Hope\:This\:Helps!!!}$

$\mathrm{-\:austint1414}$