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Let the domain be $x \in [0,h]$. We have three points, $(0,1)$, $(h/2,0)$ and $(h,0)$. How do I find the quadratic equation?

My attempt: I know that the roots are located at $x=h/2$ and $x=h$. Thus the equation must be of the form $(x-h/2)(x-h) = 0$.

Expanding out we have $x^2 - (h/2)x - hx + h^2 /2 =x^2 - (3h/2)x + h^2 /2 = 0$.

I'm not sure what to do with the y-intersection.

JoJo
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    The equation has the form $c(x-h/2)(x-h)$, or $c(2x-h)(x-h)$, where $c$ is a constant, which you can adjust with the $y$-intercept. – Bernard Dec 27 '15 at 01:04
  • "Thus the equation must be of the form $(x−h/2)(x−h) = 0$" - no, because you don't know that the coefficient of the $x^2$ term is $1$. – Alnitak Jun 14 '18 at 13:41

2 Answers2

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The equation must be of the form $a(x-\frac{h}{2})(x-h)$ with $a \neq 0$. Using the point $(0,1)$ we get that $$ a \cdot \left(-\frac{h}{2}\right) \cdot (-h) = 1, $$ so $a = 2/h^2$ (assuming that $h > 0$).

Jendrik Stelzner
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You are close. The quadratic is of the shape $y=A(x-h/2)(x-h)$, where $A$ is a constant.

Use the point $(0,1)$ to identify $A$.

André Nicolas
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  • Thank you, this worked! But I don't see the logic of multiplying by a constant, instead of, say, adding a constant? I just don't `see' it. – JoJo Dec 27 '15 at 01:13
  • Multiplying $(x-h/2)(x-h)$ by a constant does not change the roots. Adding a non-zero constant does. – André Nicolas Dec 27 '15 at 01:18