In desperate need of Math Help!!!! +rep

[John]

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

[Ray-Chill]

=] 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)

[Victoria]

(you need an account to see links)

its (-1.-9) according to this

[MasterMind0wnz]

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

[John]

ack is it negative or positive?

[zlolekim]

Its positive John

[MasterMind0wnz]

Vertex: (1,9)

[Victoria]

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

[John]

thanks guys! repping everyone who helped