# How do I find the focus of the parabola represented by y=2x^2-8x+9?

Oct 19, 2015

$\left(2 , - \frac{135}{8}\right)$

#### Explanation:

First, we must convert the equation into the vertex-form:

$\textcolor{w h i t e}{\left[1\right] X X} y = a {\left(x - h\right)}^{2} + k$

Solution:

$\left[1\right] \textcolor{w h i t e}{X X} y = 2 {x}^{2} - 8 x - 9$

$\left[2\right] \textcolor{w h i t e}{X X} y + 9 = 2 {x}^{2} - 8 x$

$\left[3\right] \textcolor{w h i t e}{X X} y + 9 = 2 \left({x}^{2} - 4 x\right)$

$\left[4\right] \textcolor{w h i t e}{X X} y + 9 + 2 \left(4\right) = 2 \left({x}^{2} - 4 x\right) + 2 \left(4\right)$

$\left[5\right] \textcolor{w h i t e}{X X} y + 9 + 8 = 2 \left({x}^{2} - 4 x + 4\right)$

$\left[6\right] \textcolor{w h i t e}{X X} y + 17 = 2 {\left(x - 2\right)}^{2}$

$\left[7\right] \textcolor{w h i t e}{X X} \textcolor{red}{y = 2 {\left(x - 2\right)}^{2} - 17}$

Now that we have the equation in vertex form, we can easily get the focus:

Focus: $\textcolor{w h i t e}{\left[1\right] X} \left(h , k + \frac{1}{4 a}\right)$

Solution:

$\left[1\right] \textcolor{w h i t e}{X X} \left(h , k + \frac{1}{4 a}\right)$

$\left[2\right] \textcolor{w h i t e}{X X} \left(\left(2\right) , \left(- 17\right) + \frac{1}{4 \left(2\right)}\right)$

$\left[3\right] \textcolor{w h i t e}{X X} \left(2 , - 17 + \frac{1}{8}\right)$

$\left[4\right] \textcolor{w h i t e}{X X} \textcolor{red}{\left(2 \text{,} - \frac{135}{8}\right)}$