Okay, so I have this math problem.

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thread starter Great Sep 10 2009 + PM | QUOTE | PERMALINK | REPORT

I'm currently studying for math.. And this problem kind of stumped me. It's a pre-calculus thing, I believe. The problem is: Write the equation of a parabola that has a vertex of (-3,3) and passes through the origin. The equation for a parabola is ax^2+bx+c...and I found that c is 0. The rest of the problem made me kind of mad. Help?

WHOAHeadshot 09/10/09 + PM | QUOTE | PERMALINK | REPORT

Great : Help? Nothx.

Great 09/10/09 + PM | QUOTE | PERMALINK | REPORT

WHOAHeadshot : Nothx. I'm being serious, and if you would not like to discuss, then please don't post.

MahnamesEric 09/10/09 + PM | QUOTE | PERMALINK | REPORT

Did you know 111,1111,111 x 111,111,111 equals 12345678987654321 or something like that. TEEHEE!

ConanEdogawa 09/10/09 + PM | QUOTE | PERMALINK | REPORT

Guys just answer her question... I'm bumping thread also :D I have no clue what it is X-X

VanLanxe 09/10/09 + PM | QUOTE | PERMALINK | REPORT

Do you want me to explain it or just give you the answer? I can solve it on my graphing calculator.

Great Sep 10 2009 + PM | QUOTE | PERMALINK | REPORT

VanLanxe : Do you want me to explain it or just give you the answer? I can solve it on my graphing calculator. I can too, but on the test tomorrow, we can't use graphing calculators, I think. If you can explain it, I would appreciate it greatly.

blast 09/10/09 + PM | QUOTE | PERMALINK | REPORT

*stays for explanation cause I'm terrible with this* :D

fullmoon 09/10/09 + PM | QUOTE | PERMALINK | REPORT

Stay online. I can explain. just give me a second

benq Sep 10 2009 + PM | QUOTE | PERMALINK | REPORT

Start with f(x)=a(x+3)^2+3 substitute x = 0, f(x) = 0 to solve for a. 0=9a+3 a=-1/3 Therefore, f(x)=-1/3(x+3)^2+3. In, ax^2+bx+c (standard form): f(x)= -1/3(x^2+6x+9)+3 = -1/3x^2-2x-3+3 = -1/3x^2-2x <---- Standard form = -x(1/3x+2) <---- Factored form

lilbabyviet 09/10/09 + PM | QUOTE | PERMALINK | REPORT

Does the problem want it in standard form? Because this is what "ax^2+bx+c" is.

NineCrimes 09/10/09 + PM | QUOTE | PERMALINK | REPORT

I used to know how to do this. I'm pissed that I've forgotten how.

KewlEH 09/10/09 + PM | QUOTE | PERMALINK | REPORT

Do you want the equation to be changed to vertex form?

Great Sep 10 2009 + PM | QUOTE | PERMALINK | REPORT

lilbabyviet : Does the problem want it in standard form? Because this is what "ax^2+bx+c" is. I don't think so. I'm pretty sure you can do it in vertex form. benq : Start with f(x)=a(x+3)^2+3 substitute x = 0, f(x) = 0 to solve for a. 0=9a+3 a=-1/3 Therefore, f(x)=-1/3(x+3)^2+3. OHH. I forgot about that because I forgot to plug in 0. I completely understand, and I thank you very much.

fullmoon 09/10/09 + PM | QUOTE | PERMALINK | REPORT

D: nevermind. Someone already explained. (lul had to look through my algebra notebook to make sure) ;~;