Given: F(x)= x^2-6x-8 , complete the square to find the vertex.
TY for your help it is much appreciated!
Think about this:
$\displaystyle f(x)=x^2-ax+b$
$\displaystyle f(x)=x^2-ax+b+c-c$ 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.
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!
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.
$\displaystyle f(x) = x^2-6x-8$
$\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)