Defining a quadratic equation from its definite integral and vertex

Factual Questions
1

As always, not homework. Just trying to expand my math skillz.

Given the following:

  • a function has the form -ax^2 + bx + c = 0 (i.e., I know it is a downward facing parabola)
  • I know the coordinates of the vertex, and the y value of that is positive
  • I know the the numerical value of the definite integral, between the two x intercepts

Can I define a and b from this information? I know that I can define c simply because I know the vertex. I’ve been trying to work this out for some time, and I can’t get it. Instinctively, I think this should be doable.

2

What do you mean you can define c? It will depend on a and b, not only on the vertex.

3

Good catch. I forgot to include that the vertex is on the y-axis. So let’s update the specifications:

  • a function has the form -ax^2 + bx + c = 0 (i.e., I know it is a downward facing parabola)
  • I know the value of the vertex, x = 0 and the y value is positive
  • I know the numerical value of the definite integral, between the two x intercepts
4

If you know the vertex, (h, k), then you can write y = –a(x–h)[sup]2[/sup] + k, where the only thing undetermined is the value of a. (This could then be multiplied out into the “-ax^2 + bx + c” form.)

The greater a is, the narrower the parabola is, and the smaller the value of the definite integral you’re talking about (= the area of the region between the parabola and the x-axis).

So the answer is yes, but the argument I’ve given isn’t a constructive proof.

On preview: The further condition you added, that the vertex be on the y-axis, isn’t necessary, but it probably makes it easier.

5

In the general case you have four unknowns, so you need four equations.

If we call the function f, the vertex (d, e) and the value of the integral g and the distance from x=d to the x-intercepts h we get.

f(d)=e
f’(d)=0
f(d-e) or f(d+e)=0
And the definite integral of f from d-e to d+e = g.

That’s four equations for the unknowns a, b, c, e.

6

If the vertex is on the y axis, that would mean b=0, doesn’t it?

7

But those aren’t four independent unknowns.

8

No, but it was easier, for me at least, to find a, b and c that way from the premises given, than to eliminate the unknowns.

And the set of equations did give valid solutions for a random vertex and integral I tried in Geogebra.

9

Oh, wait. Did you thing I meant d, e etc. were unknowns? I know I didn’t specify properly, but the unknowns in question are a, b, c and h.

10

:confused:
I thought we were dealing with three independent unknowns (a, b and c), not four.

(I assume the OP meant y = -ax^2 + bx + c, not -ax^2 + bx + c = 0).

11

Yes. I added h as the distance between the x-value of the vertex and the x intercepts because otherwise you’d have to but the whole quadratic formula into the definite integral. (Although I didn’t figure that out at the time, I just realised adding another unknown made for a cleaner set of equations for my to type into the CAS software.)

12

Adapted for clarity and universality.

We have the equation A: a x^2 + b x + c = 0 and function f(x) = a x^2 + b x + c

With a, b and c being unknown, and with the givens:
A having two solutions
a being negative
f having the vertex (d, e)
the integral of the function between the x-intercepts being g.

If we call the two solutions to A A[sub]1[/sub] and A[sub]2[/sub]

Then
f(d)=e
f’(d) = 0
And the absolute value of the definite integral of f from A[sub]1[/sub] to A[sub]2[/sub] = g.

Three unknowns, three equations.

And I think we can remove the condition of a being negative and just add that we’re comparing with the absolute value of g.

13

Thank you, everyone! That’s really helpful!

14

I worked it out like this. As said b = 0 and then c is the actual value when x = 0, i.e. the value at the y-intercept (vertex). Then the x-intercepts are at ± sqrt(c/a) (simple computation, both a and c are assumed positive) and the integral between them works out to (4c)/c * sqrt(c/a). Since this and c are known, you can solve for a.