# Vertex Equation of an inverse quadratic function.

I'm working on a graphing web tool using JSXGraph, The user should be able to draw different functions. I was able to allow the user to draw quadratic functions by creating the vertex of the function where the user clicks, and then creating another point up-right of the the vertex. Then I construct the function by obtaining $a$ through the vertex equation

$y = a(x - h)^2 + k$

I isolate $a$, which gives

$a = \frac{y-k}{x^2-2xh+h^2}$

As I have the vertex and another point, I obtain $a$, and then isolating $y$

$f(x) = ax^2-2ahx+ah^2+k$

with that, I obtain $b$ and then $c$.

The problem is I also need to allow the user to draw inverted parabolas, however I can't find a way to obtain the function given the vertex and a point. I believe i would be able to, with the vertex equation of the inverse quadratic, following a similar procedure as the one above, but I'm not even sure if that exists given the dual nature of the inverse quadratic.

So is there a vertex equation for the inverse of a quadratic function that would allow me to accomplish what I did above?

Thanks.

As you explained above, a parabola can be uniquely defined by its vertex $V=(v_x, v_y)$ and one more point $P=(p_x, p_y)$. The function term of the parabola then has the form $$y = a (x-v_x)^2 + v_y.$$ Then, $a$ can be determined by solving $$p_y = a (p_x-v_x)^2 + v_y$$ for $a$ which gives $$a = (p_y - v_y) / (p_x - v_x)^2.$$
Conversely, also the inverse quadratic function can be uniquely defined by its vertex $V=(v_x, v_y)$ and one more point $P=(p_x, p_y)$. The function term of the inverse function has the form $$y = \sqrt{\frac{x-v_x}{a}} + v_y.$$ Again, $a$ can be determined by solving $$p_y = \sqrt{\frac{p_x-v_x}{a}} + v_y$$ for $a$ which gives $$a = (p_x - v_x) / (p_y - v_y)^2 .$$ I set up an example in the JSXGraph wiki.