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quadratic has the form f(x)=ax^2+bx+c right?

so we have 9a-3b=18 or 3a-b=6 if we divide both sides by 3

but we have two unknowns so now i'm stuck maybe we should try the form f(x)=a(x-h)^2+k

I don't know much, but if I am correct, the equation must be of the form ay^2+by+c=0

As you see there is only one x intercept

you can use either form

Though I am not sure

we know the y-intercept which is (0,18) so we have f(x)=a(x-h)^2+18 now we need to find h and a

0=f(-3)=a(0-h)^2+18=ah^2+18 so we have ah^2=-18

so either way we still have two unknowns

oh i see what you are saying no y=f(x) not x=y

Now you got it

its f(x)=ax^2+bx+c or you can use f(x)=a(x-h)^2+k

we don't know anything else about the parabola?

I beg your pardon?

If you find any solution please post it here, I will have to leave now, as there is a black out here

i was just asking if we know anything else about the parabola?

go lokisan!

haha, just saw what you wrote, myininaya...

lol .

ignorant, do we know anything else about the parabola?

like the vertex or something else?

No that is all the information I have about it

do you know derivatives?

Yes sure

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

ok awesome

f'(x)=2ax+b

Forget the matrix stuff if you haven't done it. You can use the equations.

Actually I happen to have learnt that rotation part of matrix

Good