I drew out the parabola and attached it below just in case it helps. The maximum height of the parabola was given at $18$, the starting point of the parabola was given at $(0,0)$ and the ending point of the parabola (where it encounters the x-axis again) was given at $(18,0)$.
I wrote out the x coordinate to be $9$ only because the graph peaks exactly half way through but is this correct? Seems too simple.
A further question which was asked was to find the actual quadratic equation of this graph. I also tried that out with the three points $(0,0),(18,0),(9,18)$. Using the formula $y=ax^2+bx+c$ I substituted all of the values and got the answer $y=\frac{-324}{1458}x^2+\frac{5832}{1458}x+0$
(reference for how I worked this out is at https://www.youtube.com/watch?v=MMl8VHt1nU4)
Use: $$y=ax^2+bx+c$$ and substitute the equation above with three coordinates which are $(0,0)$, $(9,18)$ and $(18,0)$ and you will get three equations. $$\left(0,0\right)\::\:c=0$$ $$\left(9,18\right)\::\:81a+9b+c=18$$ $$\left(18,0\right)\::\:324a+18b+c=0$$ Substitute $c=0$ for the second and third equation $$81a+9b=18$$ $$324a+18b=0$$ Since this is an easy linear system, you should know how to solve it. So, the value is $$a=-\frac{2}{9}$$ $$b=4$$ $$c=0$$ The equation is: $$y=ax^2+bx+c$$ $$y=-\frac{2}{9}x^2+4x$$
Image reference is below: