# 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  Reply With Quote

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)  Reply With Quote

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its (-1.-9) according to this  Reply With Quote

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

5. ack is it negative or positive?  Reply With Quote

6. Its positive John   Reply With Quote

7. Vertex: (1,9)  Reply With Quote

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

9. thanks guys! repping everyone who helped  Reply With Quote

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