# How might I define a parabola in vertex form, such that…

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! :)

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

$$y=\frac{-k}{h^2}(x-h)^2+k$$

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.

• 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. :) – even87 Oct 11 at 10:55