Given $(0,0)$ origin, point $(3,2)$ and $y\text{-max} = 5$, find the parabola. I tried to shift point $(3,2)$ down to $(3,0)$ so that it can become symmetric to origin. Then the vertex would be $(5,\frac 3 2)$ then using the equation $$(x-h)^2=4p(y-k).$$

However after doing this the point $(0,0)$ does not lie within the parabola.

  • $\begingroup$ You’re assuming that the axis of symmetry is parallel to the $y$-axis. Without some assumption about this axis or other assumptions, there’s not enough information to solve the problem. $\endgroup$ – amd Mar 20 '19 at 4:29

Let $$y = -a(x-b)^2 + c$$

We know that $c=5$.

At point $(0,0)$, we have $0=-ab^2+5$, that is $$ab^2=5$$

At point $(3,2)$, we have $2=-a(3-b)^2+5$, that is $$a(3-b)^2=3$$

Solve for $a$ and $b$.

To solve for $b$, note that we have $$5(3-b)^2=3b^2$$

which is just a quadratic equation, try to take it from here.


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.