# Defining a quadratic equation from its definite integral and vertex **URL:** \ **Category:** Factual Questions\ **Created:** [May 4, 2018, 2:57pm UTC](https://boards.straightdope.com/t/defining-a-quadratic-equation-from-its-definite-integral-and-vertex/813581 "2018-05-04T14:57:35Z")\ **Posts on this page:** 14\ **Page:** 1
**Author:** ![D18](https://avatars.discourse-cdn.com/v4/letter/d/c57346/32.png) [@D18](https://boards.straightdope.com/u/D18)\ **Post date:** [May 4, 2018, 2:57pm UTC](https://boards.straightdope.com/t/defining-a-quadratic-equation-from-its-definite-integral-and-vertex/813581/1 "2018-05-04T14:57:35Z")
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. ---
**Author:** ![naita](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/naita/32/5862_2.png) [@naita](https://boards.straightdope.com/u/naita)\ **Post date:** [May 4, 2018, 3:17pm UTC](https://boards.straightdope.com/t/defining-a-quadratic-equation-from-its-definite-integral-and-vertex/813581/2 "2018-05-04T15:17:04Z")
What do you mean you can define _c_? It will depend on _a_ and _b_, not only on the vertex. ---
**Author:** ![D18](https://avatars.discourse-cdn.com/v4/letter/d/c57346/32.png) [@D18](https://boards.straightdope.com/u/D18)\ **Post date:** [May 4, 2018, 3:28pm UTC](https://boards.straightdope.com/t/defining-a-quadratic-equation-from-its-definite-integral-and-vertex/813581/3 "2018-05-04T15:28:01Z")
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 ---
**Author:** ![Thudlow\_Boink](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/thudlow_boink/32/320_2.png) [@Thudlow\_Boink](https://boards.straightdope.com/u/Thudlow_Boink)\ **Post date:** [May 4, 2018, 3:35pm UTC](https://boards.straightdope.com/t/defining-a-quadratic-equation-from-its-definite-integral-and-vertex/813581/4 "2018-05-04T15:35:16Z")
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. ---
**Author:** ![naita](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/naita/32/5862_2.png) [@naita](https://boards.straightdope.com/u/naita)\ **Post date:** [May 4, 2018, 3:50pm UTC](https://boards.straightdope.com/t/defining-a-quadratic-equation-from-its-definite-integral-and-vertex/813581/5 "2018-05-04T15:50:18Z")
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. ---
**Author:** ![scr4](https://avatars.discourse-cdn.com/v4/letter/s/59ef9b/32.png) [@scr4](https://boards.straightdope.com/u/scr4)\ **Post date:** [May 4, 2018, 3:57pm UTC](https://boards.straightdope.com/t/defining-a-quadratic-equation-from-its-definite-integral-and-vertex/813581/6 "2018-05-04T15:57:38Z")
If the vertex is on the y axis, that would mean b=0, doesn’t it? ---
**Author:** ![Thudlow\_Boink](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/thudlow_boink/32/320_2.png) [@Thudlow\_Boink](https://boards.straightdope.com/u/Thudlow_Boink)\ **Post date:** [May 4, 2018, 4:01pm UTC](https://boards.straightdope.com/t/defining-a-quadratic-equation-from-its-definite-integral-and-vertex/813581/7 "2018-05-04T16:01:08Z")
> [@naita](#): > > 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… But those aren’t four _independent_ unknowns. ---
**Author:** ![naita](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/naita/32/5862_2.png) [@naita](https://boards.straightdope.com/u/naita)\ **Post date:** [May 4, 2018, 4:09pm UTC](https://boards.straightdope.com/t/defining-a-quadratic-equation-from-its-definite-integral-and-vertex/813581/8 "2018-05-04T16:09:26Z")
> [@Thudlow\_Boink](#): > > But those aren’t four _independent_ unknowns. 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. ---
**Author:** ![naita](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/naita/32/5862_2.png) [@naita](https://boards.straightdope.com/u/naita)\ **Post date:** [May 4, 2018, 4:10pm UTC](https://boards.straightdope.com/t/defining-a-quadratic-equation-from-its-definite-integral-and-vertex/813581/9 "2018-05-04T16:10:45Z")
> [@Thudlow\_Boink](#): > > But those aren’t four _independent_ unknowns. 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. ---
**Author:** ![scr4](https://avatars.discourse-cdn.com/v4/letter/s/59ef9b/32.png) [@scr4](https://boards.straightdope.com/u/scr4)\ **Post date:** [May 4, 2018, 4:16pm UTC](https://boards.straightdope.com/t/defining-a-quadratic-equation-from-its-definite-integral-and-vertex/813581/10 "2018-05-04T16:16:18Z")
> [@naita](#): > > 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. 😕 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). ---
**Author:** ![naita](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/naita/32/5862_2.png) [@naita](https://boards.straightdope.com/u/naita)\ **Post date:** [May 4, 2018, 4:24pm UTC](https://boards.straightdope.com/t/defining-a-quadratic-equation-from-its-definite-integral-and-vertex/813581/11 "2018-05-04T16:24:53Z")
> [@scr4](#): > > 😕 > 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). 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.) ---
**Author:** ![naita](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/naita/32/5862_2.png) [@naita](https://boards.straightdope.com/u/naita)\ **Post date:** [May 4, 2018, 4:38pm UTC](https://boards.straightdope.com/t/defining-a-quadratic-equation-from-its-definite-integral-and-vertex/813581/12 "2018-05-04T16:38:30Z")
> [@naita](#): > > 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. 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. ---
**Author:** ![D18](https://avatars.discourse-cdn.com/v4/letter/d/c57346/32.png) [@D18](https://boards.straightdope.com/u/D18)\ **Post date:** [May 4, 2018, 8:25pm UTC](https://boards.straightdope.com/t/defining-a-quadratic-equation-from-its-definite-integral-and-vertex/813581/13 "2018-05-04T20:25:50Z")
Thank you, everyone! That’s really helpful! ---
**Author:** ![Hari\_Seldon](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/hari_seldon/32/5173_2.png) [@Hari\_Seldon](https://boards.straightdope.com/u/Hari_Seldon)\ **Post date:** [May 5, 2018, 12:01am UTC](https://boards.straightdope.com/t/defining-a-quadratic-equation-from-its-definite-integral-and-vertex/813581/14 "2018-05-05T00:01:50Z")
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.