What is the equation of the parabola that has a vertex at  (6, 2)  and passes through point  (3,20) ?

Jun 25, 2018

$y = 2 {\left(x - 6\right)}^{2} + 2$

Explanation:

Given:
$\textcolor{w h i t e}{\text{XXX}}$Vertex at $\left(\textcolor{red}{6} , \textcolor{b l u e}{2}\right)$, and
$\textcolor{w h i t e}{\text{XXX}}$Additional point at $\left(3 , 20\right)$

If we assume the desired parabola has a vertical axis,
then the vertex form of any such parabola is
$\textcolor{w h i t e}{\text{XXX}} y = \textcolor{g r e e n}{m} {\left(x - \textcolor{red}{a}\right)}^{2} + \textcolor{b l u e}{b}$ with vertex at $\left(\textcolor{red}{a} , \textcolor{b l u e}{b}\right)$

Therefore our desired parabola must have the vertex form
$\textcolor{w h i t e}{\text{XXX}} y = \textcolor{g r e e n}{m} {\left(x - \textcolor{red}{6}\right)}^{2} + \textcolor{b l u e}{2}$

Furthermore we know that the "additional point" $\left(x , y\right) = \left(\textcolor{m a \ge n t a}{3} , \textcolor{t e a l}{20}\right)$

Therefore
$\textcolor{w h i t e}{\text{XXX}} \textcolor{t e a l}{20} = \textcolor{g r e e n}{m} {\left(\textcolor{m a \ge n t a}{3} - \textcolor{red}{6}\right)}^{2} + \textcolor{b l u e}{2}$

$\textcolor{w h i t e}{\text{XXX}} \Rightarrow 18 = 9 \textcolor{g r e e n}{m}$

$\textcolor{w h i t e}{\text{XXX}} \Rightarrow \textcolor{g r e e n}{m} = 2$

Plugging this value back into our earier version of the desired parabola, we get
$\textcolor{w h i t e}{\text{XXX}} y = \textcolor{g r e e n}{2} {\left(x - \textcolor{red}{6}\right)}^{2} + \textcolor{b l u e}{2}$

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If the axis of symmetry is not vertical:
[1] if it is vertical a similar process can be used working with the general form $x = m {\left(y - b\right)}^{2} + a$
[2] if it is neither vertical nor horizontal, the process becomes more involved (ask as a separate question if this is the case; in general you will need to know the angle of the axis of symmetry in order to develop an answer).