# How do you write in vertex form for y=x^2 + 8x - 7 by completing the square?

Jun 13, 2015

$y = {x}^{2} + 8 x - 7 = {\left(x + 4\right)}^{2} - 23$

#### Explanation:

${x}^{2} + 8 x - 7$ is of the form $a {x}^{2} + b x + c$

with $a = 1$, $b = 8$ and $c = - 7$.

Notice that in general:

$a {\left(x + \frac{b}{2 a}\right)}^{2} = a {x}^{2} + b x + {b}^{2} / \left(4 a\right)$

In our case $\frac{b}{2 a} = \frac{8}{2} = 4$, so we want ${\left(x + 4\right)}^{2}$

${\left(x + 4\right)}^{2} = {x}^{2} + 8 x + 16$

So

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

$= {x}^{2} + 8 x + 16 - 16 - 7$

$= {\left(x + 4\right)}^{2} - 16 - 7$

$= {\left(x + 4\right)}^{2} - 23$