Given the formula for a parabola in vertex form: $y = a(x-h)^2+k$, as $h$ and $k$ are changed, the $a$ value will adjust in order to keep the left hand $x$-intercept anchored to the origin. I'm really only interested in quadrant $I$, and between $0 ≤ x,y ≤ 1$.

What I'm really looking to do is graph a curve (the type is negotiable) that rises from $(0, 0)$, peaks at some target/adjustable number near $(0.5, 1)$, and then falls, depending on where the peak is, to $(1, 0)$.

I considered using a spline for this, but this curve will drive a function in an inefficient node based system, where I need to keep the math complexity is low as possible. A parabola seems like the best fit.

Thanks! :)


1 Answer 1


You need your parabola to go through the origin; to achieve this, simply plug $\,(0,0)\,$ into $\,(x,y)\,$ and solve for $\,a.\,$

$$ \begin{align} y&=a(x-h)^2+k\\[1ex] 0&=a(0-h)^2+k\\[1ex] \end{align} $$

This yields $\,a=\frac{-k}{h^2}.\,$ Thus the parabola


will always go through the origin.

To satisfy your criteria that the origin be the left intercept, you simply need $\,h\gt0;\,$ and $\,k\gt0\,$ will ensure the parabola opens down. This should fit your requirements exactly.

  • 1
    $\begingroup$ Perfect, thank you! And in response to the (now deleted?) comment asking what the terms refer to, I've included the standard formula for future reference. I should have done that to begin with. :) $\endgroup$
    – even87
    Oct 11, 2019 at 10:55

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