# Find the parabola given two points and $y$-max (no axis of symmetry)

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.

• 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. – 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.