Published 12 March, 2009; last updated on 15 March, 2019; originally posted at http://blog.craiga.id.au/post/85955946/determining-a-parabola-with-a-vertex-and.

So I needed to figure how to come up with a formula for the following graph, given the vertex (i.e. the “turning point”) and the y-intercept.

I remembered the polynomial form from high school, but that only gave me direct control over the y-intercept, and I was able to find the vertex form formula on Wikipedia which gave me the ability to specify a vertex, but gave me no control over the y-intercept. So, after stretching my brain a bit to remember my high school maths, I came up with the following formula:

`h`and`k`are the`x`and`y`coordinates of the vertex`c`is the y-intercept

So, in the above example, we substituted `h` and `k` for 1 and 2 (the coordinates of the vertex), and `c` for 1 (the y-intercept).