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

Question

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 $y=45$.

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Answer 1

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:

Answer 2

There are various methods for finding the equation of the parabola; the methods themselves aren't particularly difficult, but the coordinate values given in the problem make the result look "ugly".

Since DMcMor worked out the "vertex form" of the equation, we will develop the "standard form" here (it isn't any "prettier"). We are given the $ \ y-$intercept of the parabola, so we know that the equation has the form $ \ y \ = \ ax^2 + bx + 10 \ $ at the start, along with the fact that $ \ a < 0 \ \ $ since the parabola "opens downward". We are further told that one of the $ \ x-$ intercepts is located at $ \ x = 7 \ \ . $

The axis of symmetry of the parabola passes through the vertex at $ \ (h \ , \ 45) \ \ . $ We can use a proportionality property to find this $ \ x-$coordinate. With the vertex form of the equation expressed as $ \ y - 45 \ = \ a·(x - h)^2 \ \ , $ having the coordinates of two points on the curve, $ \ (X_1 \ , \ Y_1) \ $ and $ \ (X_2 \ , \ Y_2) \ \ , $ allows us to write $$ \frac{|Y_2 - 45|}{|Y_1 - 45|} \ = \ \frac{|a|·(X_2 - h)^2}{|a|·(X_1 - h)^2} \ = \ \frac{ (X_2 - h)^2}{ (X_1 - h)^2} \ \ . $$ For the two known points on the parabola, we can determine $$ \frac{|10 - 45|}{|0 - 45|} \ = \ \frac{35}{45} \ = \ \frac{7}{9} \ = \ \frac{ (0 - h)^2}{ (7 - h)^2} \ \ \Rightarrow \ \ \frac{ h}{ 7 - h } \ \ = \ \ \pm \frac{\sqrt7}{3} \ \ ; $$ since the vertical distance of $ \ (0 \ , \ 10) \ $ from the vertex is smaller than for $ \ (7 \ , \ 0) \ \ , $ the vertex is a smaller horizontal distance from $ \ (0 \ , \ 10) \ \ , $ so we have $$ h \ \ = \ \ \frac{\sqrt7}{3} · ( 7 - h ) \ \ \Rightarrow \ \ \left(\frac{3 \ + \ \sqrt7}{3} \right)·h \ \ = \ \ \frac{7\sqrt7}{3} $$ $$ \Rightarrow \ \ h \ \ = \ \ \frac{7\sqrt7}{3 + \sqrt7} \ \ = \ \ \frac{7\sqrt7·(3 - \sqrt7)}{2} \ \ \approx \ \ 3.280 \ \ . $$

The axis of symmetry is midway between the $ \ x-$intercepts of a parabola, so we can calculate the second intercept as $$ h \ = \ \frac{\mathcal{X} \ + \ 7}{2} \ \ = \ \ \frac{7\sqrt7·(3 - \sqrt7)}{2} $$ $$ \Rightarrow \ \ \mathcal{X} \ \ = \ \ 2h - 7 \ \ = \ \ 7·(3\sqrt7 - 7) \ - \ 7·1 \ \ = \ \ 7·(3\sqrt7 - 8) \ \ \approx \ \ -0.439 \ \ . $$

Now that we know the zeroes of the quadratic polynomial (the $ \ x-$intercepts of the parabola), we can write the "factored form" $ \ y \ = \ a·(x - 7)·( \ x \ - \ [7·(3\sqrt7 - 8)] \ ) \ \ . $ For $ \ x = 0 \ \ , $ this must give us the $ \ y-$coordinate of the $ \ y-$intercept, thus $$ a·(0 - 7)·( \ 0 \ - \ [7·(3\sqrt7 - 8)] \ ) \ \ = \ \ 10 $$ $$ \Rightarrow \ \ a \ \ = \ \ \frac{10}{49·(3\sqrt7 - 8)} \ \ = \ \ \frac{10·(3\sqrt7 + 8)}{49·([3\sqrt7]^2 - 8^2)} \ \ = \ \ \frac{10·(3\sqrt7 + 8)}{49·(63 - 64)} $$ $$ = \ \ -\frac{10·(3\sqrt7 + 8)}{49} \ \ \approx \ \ -3.253 \ \ . $$

The middle coefficient $ \ b \ $ of the standard form can be found now by multiplying out the factored form, or by using the relation for the vertex $ \ h \ = \ -\frac{b}{2a} \ \ , $ from which we obtain $$ b \ \ = \ \ -2·a·h \ \ = \ \ -2·\left(-\frac{10·[3\sqrt7 + 8]}{49} \right)·\left(\frac{7\sqrt7·(3 - \sqrt7)}{2} \right) $$ $$ = \ \ \frac{10· \sqrt7}{7} · (3\sqrt7 + 8) · (3 - \sqrt7) \ \ = \ \ \frac{10· (7 + 3\sqrt7)}{7} \ \ \approx \ \ 21.339 \ \ . $$

The standard form of the parabolic equation is therefore $$ y \ \ = \ \ \left(-\frac{10·(3\sqrt7 + 8)}{49} \right)·x^2 \ + \ \left(\frac{10· (7 + 3\sqrt7)}{7} \right)·x \ + \ 10 $$ $$ \approx \ \ -3.253·x^2 \ + \ 21.339·x \ + \ 10 \ \ . $$ (Told you it wouldn't be pretty...)

Inserting $ \ x = h \ $ and $ \ x = 7 \ $ produces the given $ \ y-$coordinates of $ \ 45 \ $ and $ \ 0 \ \ , $ respectively.