# What is the vertex form of 5y = -9x^2-4x+2?

Aug 28, 2017

$y = - \frac{9}{5} {\left(x + \frac{2}{9}\right)}^{2} + \frac{22}{45}$

#### Explanation:

A quadratic function of the form $y = a {x}^{2} + b x + c$ in vertex form is given by:

$y = a {\left(x - h\right)}^{2} + k$ where $\left(h , k\right)$ is the vertex of the parabola.

The vertex is the point at which the parabola intersects its axis of symmetry. The axis of symmetry occurs where $x = \frac{- b}{2 a}$

In our example: $5 y = - 9 {x}^{2} - 4 x + 2$

$\therefore y = - \frac{9}{5} {x}^{2} - \frac{4}{5} x + \frac{2}{5}$

Hence, $a = - \frac{9}{5} , b = - \frac{4}{5} , c = \frac{2}{5}$

At the axis of symmetry $x = \frac{- \left(- \frac{4}{5}\right)}{2 \cdot \left(- \frac{9}{5}\right)}$

$= - \frac{4}{2 \cdot 9} = - \frac{2}{9} \approx - 0.222$

(This is the $x -$component of the vertex, $h$)

So, $y$ at the vertex is $y \left(- \frac{2}{9}\right)$

$= - \frac{9}{5} {\left(- \frac{2}{9}\right)}^{2} - \frac{4}{5} \left(- \frac{2}{9}\right) + \frac{2}{5}$

$= - \frac{4}{5 \cdot 9} + \frac{4 \cdot 2}{5 \cdot 9} + \frac{2}{5}$

$= \frac{- 4 + 8 + 18}{45} = \frac{22}{45} \approx 0.489$

(This is the $y -$component of the vertex, $k$)

Hence, the quadratic in vertex form is:

$y = - \frac{9}{5} {\left(x + \frac{2}{9}\right)}^{2} + \frac{22}{45}$

We can see the vertex on the graph of $y$ below.

graph{-9/5x^2-4/5x+2/5 [-3.592, 3.336, -2.463, 1.002]}