How do I find the equation for a parabola when given 2 points and the maximum values?

I need an equation for a parabola that passes through $2$ points and has a specific maximum value. The points are $(0, 10)$ and $(7, 0)$, and the maximum is $y=45$.

We start with "vertex form": $y = a(x - h)^2 + k$, where the point $(h,k)$ is the vertex. This tells us that $k = 45$, because the maximum occurs at $y = 45$. Now, we have $$y = a(x - h)^2 + 45$$ and two additional points that the graph has to pass through: $(0,10)$ and $(7,0)$. We can use those two points to find the two remaining parameters, $a$ and $h$: \begin{align} 10 &= ah^{2} + 45\tag{1}\\ 0 &= a(7-h)^{2} + 45.\tag{2} \end{align} Now, solving for $a$ in $(1)$ gives us $$a = -\frac{35}{h^{2}}.\tag{3}$$ Substituting $(3)$ into $(2)$ gives us $$0 = -\frac{35}{h^{2}}(7 - h)^{2} + 45 = -\frac{35}{h^{2}}(49 - 14h + h^{2}) + 45.\tag{4}$$ Simplifying $(4)$ gives us \begin{alignat}{2} \frac{9}{7}h^{2} = 49 - 14h + h^{2}&\implies\frac{2}{7}h^{2}+14h - 49 &&= 0\\ &\implies2h^{2} + 98h - 343 &&= 0.\tag{5} \end{alignat} We can solve $(5)$ using the quadratic formula: $$h = \frac{-98 \pm \sqrt{(98)^2+(4)(2)(343)}}{4} = \frac{1}{2}\left(-49 \pm 21\sqrt{7}\right) \approx −52.28, \ 3.28.$$ Now, because $h$ is the $x$-coordinate of the vertex, we need to take $h>0$ based on the provided graph, so we have $$h = \frac{1}{2}\left(-49 + 21\sqrt{7}\right) = \frac{7}{2}\left(3\sqrt{7} - 7\right).$$ Lastly, then, we use this value of $h$ along with $(3)$ to get $$a = -\frac{35}{h^{2}} = -\frac{35}{\left(\frac{7}{2}\left(3\sqrt{7} - 7\right)\right)^{2}} = -\frac{10}{49}\left(3\sqrt{7}+8\right).$$ Altogether, then, your equation is $$\boxed{y = -\frac{10}{49}\left(3\sqrt{7}+8\right)\left(x - \frac{7}{2}\left(3\sqrt{7} - 7\right)\right)^{2} + 45}$$ Desmos confirms that this is correct: