Question

Can someone help and or explain how to find the vertex of the parabola by completing the square?
y=x^2+8x-18

Answer 1

\[\Large\rm y=\color{orangered}{x^2+8x}\qquad\qquad-18\]Hey Haily c:
Ignore the -18 for now, we don't care about that guy.
We want to complete the square on these x's.

More generally, we think of it like this:\[\Large\rm y=\color{orangered}{x^2+bx}\]We take half of our b term, and square it.
\[\Large\rm \left(\frac{b}{2}\right)^2\]That's what will complete the square for us.
We'll add AND subtract that value from the right side.\[\Large\rm y=\color{orangered}{x^2+bx+\left(\frac{b}{2}\right)^2}-\left(\frac{b}{2}\right)^2\]The orange part condenses down to a perfect square,\[\Large\rm y=\color{orangered}{\left(x+\frac{b}{2}\right)^2}-\left(\frac{b}{2}\right)^2\]

Answer 2

Maybe a confusing explanation ^
Let's just jump into it and see how it goes.
So we'll take half of the b coefficient, half of 8 is 4, yah?
And square it.
\(\Large\rm \left(\dfrac{8}{2}\right)^2=16\)
This is the value that will complete the square for us.

Answer 3

\[\Large\rm y=\color{orangered}{x^2+8x+16}\qquad\qquad-18\]So we want to add 16 to complete our square.
But we can't just throw in a 16 willy nilly like that.
We have to balance it by subtracting 16 at the same time.\[\Large\rm y=\color{orangered}{x^2+8x+16}-16-18\]

Answer 4

Then the orange part will condense down, that is our perfect square.
What do you think Hailey? :d you with me?

Answer 5

Also, if you just want the vertex and you weren't forced to complete the square.

You would just do -b/2a

That would the x coordinate of your vertex.

Plug that back into x^2+8x-18 for the y value.

otherwise follow @zepdrix

Answer 6

Im still reading over the explanations so just a few more seconds and ill be with you.

Answer 7

Okay i think i got it for here thanks:)

Answer 8

Got it from there? c: k