Can't solve quadratic function

Question

I'm trying to get a simple quadratic function of the form $y=ax^2+bx+c$ which goes through the following points:

$$(0;0) \;(\frac d2;2)\; (d;0)$$

The solution I've calculated $3$ times on paper has always been $y=\frac8d- \frac{8x^2}{d^2}$, but this can't be true, for an arbitrary $d$, putting $x$ as $0$ should result in $0$, but $\frac8{10}-8\times{\frac0{100}}$ is $\frac 8{10}$, and not $0$.

Neither does input $d$ result in $0 \to y=\frac 8d$

Am I just approaching this problem incorrectly or am I daft?

Answer 1

The key bit to notice is that you've been given the roots of such a quadratic. Namely the roots are ${x=0}$ and ${x=d}$. This must mean your quadratic can be written of the form

$${y=Ax(x-d)}$$

(since any polynomial can be factorised over it's roots) where ${A}$ is a constant to be determined.

If we plug in ${x=\frac{d}{2}}$, want the result to be ${2}$. As an equation, that is

$${A\frac{d}{2}\left(\frac{d}{2}-d\right)=2}$$

We can then rearrange for ${A}$ to get

$${A=\frac{2}{\frac{d}{2}\left(\frac{d}{2}-d\right)}=\frac{-8}{d^2}}$$

Hence

$${y=\frac{-8}{d^2}x(x-d)=-\frac{8}{d^2}x^2 + \frac{8}{d}}$$

Answer 2

We have that

• passing through $(0,0) \implies c=0$

• passing through $(d/2,2) \implies 2=a\frac{d^2}4+b\frac{d}2$

• passing through $(d,0) \implies 0=ad^2+bd \implies a=-b\frac 1 d$

then from the third and the second we obtain

• $2=-b\frac d 4+b\frac d 2 \implies b=\frac 8 d \land a=-\frac 8{d^2}$

that is

$$y=-\frac 8{d^2}x^2+\frac 8 d x$$

Answer 3

Lagrange's interpolation formula instantly yields $$y(x)=2\,\frac{x(x-d)}{\frac d2\bigl(\frac d2-d\bigr)}.$$

Answer 4

By subtitution,

$$\begin{cases}c=0,\\a\dfrac{d^2}4+b\dfrac d2+c=2,\\ad^2+bd+c=0.\end{cases}$$

Dropping $c$ and subtracting twice the second from the third,

$$a\frac{d^2}2=-4\to a=-\frac 8{d^2}$$ and

$$b=-ad=\frac8d.$$

The expression is

$$\frac8dx-\frac8{d^2}x^2,$$ you missed an $x$ factor.

Answer 5

You could also apply the "vertex form" of the equation for a parabola, $ \ y \ = \ a·(x-h)^2 + k \ \ . $ Because of the curve's symmetry, when the parabola has $ \ x-$intercepts, its axis of symmetry passes midway between them. With those intercepts given as $ \ x = 0 \ $ and $ \ x = d \ \ , $ the symmetry axis is $ \ x \ = \ \frac{d}{2} \ \ . $ So the third given point is the vertex.

The equation in this form is then $ \ y \ = \ a·\left(x - \frac{d}{2} \right)^2 + 2 \ \ . $ As the intercepts are at a distance $ \ \frac{d}{2} \ $ from the symmetry axis, we can determine the remaining coefficient from $$ a·\left( \frac{d}{2} \right)^2 + 2 \ \ = \ \ 0 \ \ \Rightarrow \ \ a \ \ = \ \ \frac{-2}{d^2 / 4} \ \ = \ \ -\frac{8}{d^2} \ \ . $$ So the quadratic function in question is $$ \ \left(-\frac{8}{d^2}\right) · \left(x - \frac{d}{2} \right)^2 + 2 \ \ = \ \ -\frac{8}{d^2} x^2 \ - \ \frac{8}{d^2} · (-dx) \ - \ \frac{8}{d^2} · \left(\frac{d^2}{4}\right) \ + \ 2 $$ $$ = \ \ -\frac{8·x^2}{d^2} \ + \ \frac{8·x}{d} \ - \ 2 \ + \ 2 \ \ = \ \ -\frac{8·x^2}{d^2} \ + \ \frac{8·x}{d} \ \ . $$

(You weren't "daft", you just stopped seeing your omission -- just about all of us have done that at one time or another [or one time after another...]. Sometimes the best thing to do is drop the effort, go do something else for a while (or get some sleep), and come back later for another look.)