# Thread: Complete the square to find the vertex

1. ## Complete the square to find the vertex

Given: F(x)= x^2-6x-8 , complete the square to find the vertex.

TY for your help it is much appreciated!

2. ## Re: Complete the square to find the vertex

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Originally Posted by duBxStep
Given: F(x)= x^2-6x-8 , complete the square to find the vertex.

TY for your help it is much appreciated!
Have you tried completing the square?

3. ## Re: Complete the square to find the vertex

Well, I looked up how to do it on google and this is where I got:

y=a(x-h)^2+K
Y=1(x-h)^2+k
Y+8=x^2-6x
y+8-3=x^2-6x-3

but I have no idea where to go from there, and im not even sure if i'm on the right track.

4. ## Re: Complete the square to find the vertex

Originally Posted by duBxStep
Given: F(x)= x^2-6x-8 , complete the square to find the vertex.

TY for your help it is much appreciated!
Have you been taught how to complete the square? What do you class notes and textbook have to say about it?

5. ## Re: Complete the square to find the vertex

$\displaystyle f(x)=x^2-ax+b$
$\displaystyle f(x)=x^2-ax+b+c-c$ is the same thing.
$\displaystyle f(x)=(x^2-ax+b+c)-c$ is also the same thing.

So what you need to figure out is what number added to itself makes -6. That is: $\displaystyle x+x=-6$ and if you square it what number do you get? $\displaystyle x^2=?$.

What ever that ? is you need to make the -8 in your equation equal that by adding and subtracting the some number to it. and splitting it up like above.

6. ## Re: Complete the square to find the vertex

Originally Posted by takatok
$\displaystyle f(x)=x^2-ax+b$
$\displaystyle f(x)=x^2-ax+b+c-c$ is the same thing.
$\displaystyle f(x)=(x^2-ax+b+c)-c$ is also the same thing.

So what you need to figure out is what number added to itself makes -6( $\displaystyle x+x=-6$ ) and if you square it what number do you get? $\displaystyle x^2=?$.

What ever that ? is you need to make the -8 in your equation equal that by adding and subtracting the some number to it. and splitting it up like above.
I thought that the formula was $\displaystyle f(x)=(ax^2+bx+c)$? How come the A in the given formula has shifted over to the right one place?

Okay, so -6(x+x=-6) x=1/2 for -6 to = -6 right? square it and you get 1/4.

So I need to make 1/4 = to -8? Man im confused. I know I've done this before I just forgot how to do it!

7. ## Re: Complete the square to find the vertex

So here I am now, let me know if I'm on the right track:

$\displaystyle Y=x^2-6x-8$

$\displaystyle Y+8+9=x^2-6x+9$

$\displaystyle Y+17=x^2-6x+9$

$\displaystyle Y=(x-3)^2-17$

8. ## Re: Complete the square to find the vertex

You've got it exactly. My original post was spaced badly. I corrected. Your method you just did is essentially the same as what I described.

So now from your earlier formula
$\displaystyle y=a(x-h)^2+K$

a=1, h=3, and k=-17. Fill that in with whatever you were working on.

9. ## Re: Complete the square to find the vertex

$\displaystyle f(x) = x^2-6x-8$
$\displaystyle (x-3)^2 = 8$ *Add the square of 3 to both sides; obtained by halving middle term coefficient -6x = -3

$\displaystyle (x-3)^2 = 8 + (-3^2)$
$\displaystyle (x-3)^2 = 17$
$\displaystyle x = 3 \pm\sqrt{17}$

Hence, coordinates of vertex is (3, -17)