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

Question

3733 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-06-13T22:20:34

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

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

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

Notice that in general:

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

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

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

So

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

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

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

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

How do you write in vertex form for y=x^2 + 8x - 7y=x2+8x−7y=x^2 + 8x - 7 by completing the square?