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

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Answer 1 · 2015-06-13T22:20:34

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

$x^2+8x-7$ is of the form $ax^2+bx+c$

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

Notice that in general:

$a(x+b/(2a))^2 = ax^2+bx+b^2/(4a)$

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

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

So

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

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

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

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

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