Math Problem....HELP!?!?!?!

[STBonecrusher21]

Ok, I know this really isn't that hard. But I've been doing frickin math problems all day, and can't find it in myself to do this one.

The vertex is (3,3). The y-intercept is (0,-6).

Answer needs to be in f(x)= ax^2 + bx +c form.

First one who gets it will receive something special from yours truly.

[MLSKINS]

ES, the new homework center. llam.

I would help you but I am done with math, FOREVER!!! When I see that I just blackout.

[ixcuincle]

What's a vertex?

I helped the last time someone had one of those math questions, but this is the quadratic graphing ****. If it's not a straight line, I have no clue how to graph fancy-schmancy functions. Someone else more trained in quadratics has to do this.

[Mondizzle]

How many people are gonna come in here, get embarrassed, then leave in shame? Just me? Okay

[skinsfan_1215]

Haha I haven't had a math class since I passed calculus last year... now I take out my calculator for anything. Anything with letters in it, don't ask me.

[STBonecrusher21]

What's a vertex?

There's a graph with it too. But I can't post it. It's like a curved line that is opening upwards, and the vertex is the point at which the curved line comes to it's peak.



Like that, but only one line.

[ixcuincle]

Hopefully this helps. There's a bunch of junk that I can no longer explain, because I haven't done this **** in over 7 or 8 years. When I did do it I got a C because I didn't understand a word of this.

http://en.wikipedia.org/wiki/Quadratic_function

Good luck.

[Slacky McSlackAss]

c is equal to -6.

I could solve it if I had one more equation.

[STBonecrusher21]

I hate my teacher. She'll do the simplest version of these problems, then expect us to do ones with like 9 and 10 steps to solve it.

I'm bout to burn down her house.

[STBonecrusher21]

Ok, here's some help.

f(x) = a(x-h)^2 + k Where (h,k) is the vertex.

[ixcuincle]

f(x) = a(x-3)^2 + 3

What the heck is "a"? Dumb vague variable.

I would assume one of the numbers from the y-intercept needs to be plugged in. Then you find y, x, and then go for A. Then you fill all that mumbo jumbo into the top equation in OP except for X and Y. That sounds like a solution.

Too lazy to do that so see what happens.

This is basic high school math and I'm struggling with this

[difleha]

I don't know...I'm still taking Calculus 1 though.

EDIT: Okay, you're gonna have to find standard form for y = a(x-3)^2 + 3. I would take away the a just to make it a little clearer.

[STBonecrusher21]

First one who gets it will receive something special from yours truly.

I mean I posted that and figured people would bust their *** to try and figure it out.

I guess not. I guess no one wants that special suprise.

[ixcuincle]

Can't you just plug those two numbers into your TI-83+ and use QuadReg or something? One of you who has had training with the graphing calculator, back me up on this

http://www.prenhall.com/divisions/es.../quad-reg.html

Edit : QuadReg keeps getting error. I'm guessing the two coordinates given are not enough to graph a parabola. I give up.

[DCSaints_fan]

Since you're given the vertex, you should be able to figure out one more point from the y-intercept, because parabolas are symmetric. (Hint: What will the x-value be when y is -6, on the other side of the parabola? )

Now you have three equations, and three free variables, which you should be able to solve.