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

2. =] Awesome! We just learned about this.
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. (you need an account to see links)

its (-1.-9) according to this

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

5. ack is it negative or positive?

6. Its positive John

7. Vertex: (1,9)

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

9. thanks guys! repping everyone who helped

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