anonymous
  • anonymous
Find the quadratic equation from the following data The intercepts are (-3,0) and (0,18)
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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myininaya
  • myininaya
quadratic has the form f(x)=ax^2+bx+c right?
myininaya
  • 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
  • myininaya
so we have 9a-3b=18 or 3a-b=6 if we divide both sides by 3

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myininaya
  • 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
  • anonymous
I don't know much, but if I am correct, the equation must be of the form ay^2+by+c=0
anonymous
  • anonymous
As you see there is only one x intercept
myininaya
  • myininaya
you can use either form
anonymous
  • 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
  • anonymous
Though I am not sure
myininaya
  • 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
  • myininaya
0=f(-3)=a(0-h)^2+18=ah^2+18 so we have ah^2=-18
myininaya
  • myininaya
so either way we still have two unknowns
myininaya
  • myininaya
oh i see what you are saying no y=f(x) not x=y
anonymous
  • anonymous
Now you got it
myininaya
  • myininaya
its f(x)=ax^2+bx+c or you can use f(x)=a(x-h)^2+k
myininaya
  • myininaya
we don't know anything else about the parabola?
anonymous
  • anonymous
I beg your pardon?
anonymous
  • anonymous
If you find any solution please post it here, I will have to leave now, as there is a black out here
myininaya
  • myininaya
i was just asking if we know anything else about the parabola?
myininaya
  • myininaya
go lokisan!
anonymous
  • 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
  • anonymous
haha, just saw what you wrote, myininaya...
myininaya
  • myininaya
lol .
anonymous
  • 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
  • myininaya
ignorant, do we know anything else about the parabola?
myininaya
  • myininaya
like the vertex or something else?
anonymous
  • anonymous
No that is all the information I have about it
myininaya
  • 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
  • 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
  • myininaya
do you know derivatives?
anonymous
  • anonymous
Yes sure
anonymous
  • anonymous
http://en.wikipedia.org/wiki/Rotation_matrix
myininaya
  • myininaya
ok awesome
myininaya
  • myininaya
f'(x)=2ax+b
anonymous
  • anonymous
Forget the matrix stuff if you haven't done it. You can use the equations.
anonymous
  • anonymous
Actually I happen to have learnt that rotation part of matrix
anonymous
  • anonymous
Good
anonymous
  • 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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