I am having trouble finding the vertex of the function: x^2+6x-16=0

Question

psymon · Answer

Well, these quadratics come in the form of ax^2 + bx + c. 
If you want the vertex, we first need to do (-b/2a). So in your case b is 6 and a is just 1. 
So once you get that we can continue :3

bruno102 · Answer

I have been doing the -b/2a and I don't seem to be getting the right answer. 
I had -6/2(1) at this point. Is that correct?

psymon · Answer

Yeah, which would give you -3 for an x-coordinate. Now you just replace x with -3 in your function and see what you get.

bruno102 · Answer

Would it give me -9+(-18)-16? This is where I get confused.

psymon · Answer

$$(-3)^{2}+6(-3)-16$$
This becomes:
9 - 18 - 16.
You weren't squaring the negative part of -3 :P

bruno102 · Answer

Oh, so does that equal -25? So the vertex would be -3, -25?

psymon · Answer

Correct :3

bruno102 · Answer

Ok thank you!

psymon · Answer

Yepyep ^_^