Equation of parabola through these three points.

J

jwpaine

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I have three points. The point of origin (0,0), and two points symmetrical over the y axis, lets say (-6,3) and (6,3)

Can someone help me to do the steps in finding a quadratic that creates a parabolic curve through these three points? I know I will probably have to create and solve a system of equations.

Thanks a ton.
 
Have you a formula for a parabola? Something like (y-k)^2 = 4p(x-h)? If you have such a thing, what is it, exactly? What is the point (h,k)? Hint: V-----.
 
No.
I want to find an equation for a quadratic, containing those three points.
 
Oh, well, in that case...


Have you a formula for a parabola? Something like (y-k)^2 = 4p(x-h)? If you have such a thing, what is it, exactly? What is the point (h,k)? Hint: V-----.
 
If you recognize a particular type of "quadratic", you can exploit it to your advantage.
 
I got it.

y = ax^2 + bx + c and plug in each x and y value so that I can solve the system of three equations for the three coefficients, a,b and c.
 
jwpaine said:
I got it. y = ax^2 + bx + c and plug in each x and y value so that I can solve the system of three equations for the three coefficients, a,b and c.
Click to expand...
Yes. The other suggestion would work also, however, especially since you've been provided with the vertex, (h, k).

But your method is probably at least as quick, and more-likely what is expected by the instructor. :wink:

Eliz.
 
I have to go with, "Give me a break." on that one, stapel. (A friendly one, of course.)

If the problem statement identified the curve as a parabola, why should the student be expected to ignore that information?

Note: The general method may be "probably at least as quick" ONLY because the vertex is at the Origin. Start moving off the Origin, and you've another matter entirely.

Note to student: Do NOT get caught in the trap of thinking, believing, or practicing only one way to do things. Learn every way you can and learn the advantages and disadvantages of each. Do the same problem multiple ways, just to convince youself the answer is unique and that you remember what it is you are doing.
 
Thanks tkhunny, I understand what your saying... it's just I have no idea how I would solve this, the way you described.

I am not familiar with the form: (y-k)^2 = 4p(x-h)

I know general form: ax^2+bx+c and standard form a(x - h)2 + k after completing the square.


can you please explain the form: (y-k)^2 = 4p(x-h)
 
jwpaine said:
can you please explain the form: (y-k)^2 = 4p(x-h)
Click to expand...
You'll probably meet this form in calculus, which, if I understand correctly, you haven't yet taken. In algebra, the "y = ax<sup>2</sup> + bx + c" form seems to be much more common.

You'll occasionally see the "y - k = a(x - h)<sup>2</sup>" form in algebra, where the vertex is (h, k). But this sort of "find the equation" exercise almost never includes the vertex, which is why the texts and examples probably used your method.

Eliz.
 
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