View Full Version : 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
Ray-Chill
01-17-2012, 12:02 AM
=] 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
01-17-2012, 12:06 AM
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its (-1.-9) according to this
MasterMind0wnz
01-17-2012, 12:07 AM
I have something that will help you greatly if you don't already know it. PM Me.
ack is it negative or positive?
zlolekim
01-17-2012, 12:14 AM
Its positive John :)
MasterMind0wnz
01-17-2012, 12:14 AM
Vertex: (1,9)
Victoria
01-17-2012, 12:16 AM
sorry for giving you the wrong answer ^_^ i just googled it :P seemed legit!
thanks guys! repping everyone who helped
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