1
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ Thank you for the answer Alfred, back in that time I had already solved it, but forgot to thank you in the rush. $\endgroup$ – Zerjack Mar 16 '15 at 20:32

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.