# What is the equation of the parabola that has a vertex at  (14, -9)  and passes through point  (0, 2) ?

Aug 25, 2016

$y = \frac{11}{196} {\left(x - 14\right)}^{2} - 9$

#### Explanation:

The equation of a parabola in $\textcolor{b l u e}{\text{vertex form}}$ is

$\textcolor{red}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{y = a {\left(x - h\right)}^{2} + k} \textcolor{w h i t e}{\frac{a}{a}} |}}}$
where (h ,k) are the coordinates of the vertex and a, is a constant.

here h = 14 and k = - 9, so we can write a partial equation

$y = a {\left(x - 14\right)}^{2} - 9$

To find a, substitute the coordinates of (0 ,2) a point on the parabola, into the partial equation.

$\Rightarrow a {\left(0 - 14\right)}^{2} - 9 = 2 \Rightarrow 196 a = 11 \Rightarrow a = \frac{11}{196}$

$\Rightarrow y = \frac{11}{196} {\left(x - 14\right)}^{2} - 9 \text{ is equation in vertex form}$

The equation may be expressed in $\textcolor{b l u e}{\text{standard form}}$

That is $y = a {x}^{2} + b x + c$ by distributing the bracket and simplifying.

$\Rightarrow y = \frac{11}{196} \left({x}^{2} - 28 x + 196\right) - 9 = \frac{11}{196} {x}^{2} - \frac{11}{7} x + 2$

graph{11/196(x-14)^2-9 [-20, 20, -10, 10]}