Find the quadratic equation from the following data
The intercepts are (-3,0) and (0,18)

Question

myininaya · Answer

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

myininaya · Answer

we are given that those two points lie on the quatric so we have f(-3)=a(-3)^2+b(-3)-c=0  and we have f(0)=c=18 so f(-3)=9a-3b-18=0

myininaya · Answer

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

myininaya · Answer

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

anonymous · Answer

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

anonymous · Answer

As you see there is only one x intercept

myininaya · Answer

you can use either form

anonymous · Answer

I don't think so, as you see while you use the x form, you are putting the value of x in the equation, and that I think will not work

anonymous · Answer

Though I am not sure

myininaya · Answer

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

myininaya · Answer

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

myininaya · Answer

so either way we still have two unknowns

myininaya · Answer

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

anonymous · Answer

Now you got it

myininaya · Answer

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

myininaya · Answer

we don't know anything else about the parabola?

anonymous · Answer

I beg your pardon?

anonymous · Answer

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

myininaya · Answer

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

myininaya · Answer

go lokisan!

anonymous · Answer

Iam, I may be able to look at it a bit later.  My first impressions of it are to find the slope of the line that joins the two points, use that to perform a rotation back to the x-axis, solve the equation normally and then transform back.  Or you could transform the general equation by the same angle and solve straight from the points.

Like I said, this is a cursory glance.  I don't have time available at the moment to work through it properly, but I will when I can...promise.

anonymous · Answer

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

myininaya · Answer

lol .

anonymous · Answer

And I forgot to say, you may want to translate your points to the right 3 units before rotating...so your points would be (-3,0) --> (0,0) and (0,18) --> (3,18).

Hope this is making sense.

myininaya · Answer

ignorant, do we know anything else about the parabola?

myininaya · Answer

like the vertex or something else?

anonymous · Answer

No that is all the information I have about it

myininaya · Answer

can you make sense out what lokisan is saying?  it looks like he has a way to find the equation of the parabola

anonymous · Answer

Yes, I can get the idea of what he is saying. And I am trying to work it out on pen and paper. 
If you have any more clue, please post it

myininaya · Answer

do you know derivatives?

anonymous · Answer

Yes sure

anonymous · Answer

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

myininaya · Answer

ok awesome

myininaya · Answer

f'(x)=2ax+b

anonymous · Answer

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

anonymous · Answer

Actually I happen to have learnt that rotation part of matrix

anonymous · Answer

Good

anonymous · Answer

Your problem is:

(3,18)=R(x,y) where the ( , ) are vectors and R is the rotation matrix.

You need to invert R to find (x,y)=R^{-1}(3,18).

You'll then have two points, (0,0) and (x,0) (since both points will now be on the x-axis).

I'm rushing through this...I've just thought of a 'better' method, but I'm rushed...if I can sort it out later, I will.