What is the vertex form of $y = x^{2} + 8 x - 7$ $y=x^2+8x-7$ ?

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Answer 1 · 2016-08-19T07:17:23

$y = {(x + 4)}^{2} - 23$ $y=(x+4)^2-23$

Given -

$y = x^{2} + 8 x - 7$ $y=x^2+8x-7$

The vertex form of the equation is -

$y = a {(x - h)}^{2} + k$ $y=a(x-h)^2+k$

Where

$a$ $a$ is the coefficient of $x^{2}$ $x^2$
$h$ $h$ is the $x$ $x$ coordinate of thevertex
$k$ $k$ is the $y$ $y$ coordinate of the vertex

Vertex-

$x = \frac{- b}{2 a} = \frac{- 8}{2} = - 4$ $x=(-b)/(2a)=(-8)/2=-4$

At $x = - 4$ $x=-4$

$y = {(- 4)}^{2} + 8 (- 4) - 7$ $y=(-4)^2+8(-4)-7$
$y = 16 - 32 - 7 = - 23$ $y=16-32-7=-23$

Then-

$a = 1$ $a=1$
$h = - 4$ $h=-4$
$k = - 23$ $k=-23$

Plug in the values in the formula

$y = a {(x - h)}^{2} + k$ $y=a(x-h)^2+k$

$y = {(x + 4)}^{2} - 23$ $y=(x+4)^2-23$

What is the vertex form of y=x2+8x−7y=x^2+8x-7?