# How do you find the equation of the parabola whose vertex is at (1,-2), and that passes through the point (4,16)?

Sep 7, 2015

$y = 2 {x}^{2} - 4 x$
$\textcolor{w h i t e}{\text{XXXXXXXX}}$(see note below)

#### Explanation:

Assuming the parabola is a function in standard position
with vertex at $\left(1 , - 2\right)$ and another point at $\left(4 , 16\right)$

The general vertex form is
$\textcolor{w h i t e}{\text{XXX")y=m(x-a)^2+bcolor(white)("XXX}}$ with vertex at $\left(a , b\right)$

For the given case;
$\textcolor{w h i t e}{\text{XXX}} y = m {\left(x - 1\right)}^{2} - 2$
at $\left(4 , 16\right)$
$\textcolor{w h i t e}{\text{XXX}} 16 = m {\left(3\right)}^{2} - 2 = 9 m - 2$
$\rightarrow \textcolor{w h i t e}{\text{XXX}} m = 2$

Therefore the equation is
$\textcolor{w h i t e}{\text{XXX}} y = 2 {\left(x - 1\right)}^{2} - 2$
$\textcolor{w h i t e}{\text{XXXX}} = 2 {x}^{2} - 4 x$

Note: the general vertex form assumes the parabola is in standard position (e.g. with a vertical axis of symmetry). Infinitely many alternate solutions are possible if this is not the case.
Three such parabolas are shown below: