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

- anonymous

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

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- schrodinger

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- myininaya

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

- myininaya

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

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

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## More answers

- myininaya

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

- anonymous

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

- anonymous

As you see there is only one x intercept

- myininaya

you can use either form

- anonymous

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

Though I am not sure

- myininaya

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

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

- myininaya

so either way we still have two unknowns

- myininaya

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

- anonymous

Now you got it

- myininaya

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

- myininaya

we don't know anything else about the parabola?

- anonymous

I beg your pardon?

- anonymous

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

- myininaya

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

- myininaya

go lokisan!

- anonymous

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

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

- myininaya

lol .

- anonymous

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

ignorant, do we know anything else about the parabola?

- myininaya

like the vertex or something else?

- anonymous

No that is all the information I have about it

- myininaya

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

- anonymous

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

do you know derivatives?

- anonymous

Yes sure

- anonymous

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

- myininaya

ok awesome

- myininaya

f'(x)=2ax+b

- anonymous

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

- anonymous

Actually I happen to have learnt that rotation part of matrix

- anonymous

Good

- anonymous

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.

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