# Thread: In desperate need of Math Help!!!! +rep

1. ## In desperate need of Math Help!!!! +rep

i have a few questions that are really pissing me off. if anyone could walk me through them i would greatly appreciate it.

also i will +rep anyone who can help.

Find the Vertex of the function (put the function in vertex for first): f(x)=8+2x-x^2

xD
Firstly, put your equation into the format ax^2+bx+c=0. I'm assuming you know the quadratic formula to find the x intercepts,
x = (-b +or- SQRT(b^2-4ac))/2a

The vertex will lie on the axis of symmetry of the parabola, which is halfway between the x-intercepts ie -b/2a. Use that to work out x, and then substitute it back into the quadratic to find the y value.

Eg1. y=-x^2+2x+8 here a=-1, b=2 and c=8
So -b/2a = -2/2(-1)
= 1
Then substitute x=1 into the quadratic
y=-1+2+8
=9
Therefore the vertex is (1, 9)

3. I have something that will help you greatly if you don't already know it. PM Me.

4. ack is it negative or positive?

5. Its positive John

6. sorry for giving you the wrong answer ^_^ i just googled it seemed legit!

7. thanks guys! repping everyone who helped

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