# How do you find the axis of symmetry, graph and find the maximum or minimum value of the function y = 2x^2 - 4x -3?

Axis of symmetry$\textcolor{b l u e}{\text{ } x = 1}$
Minimum value of the function $\textcolor{b l u e}{= - 5}$
See the explanation for the graph

#### Explanation:

The solution:

To find the Axis of symmetry you need to solve for the Vertex $\left(h , k\right)$

Formula for the vertex:
$h = \frac{- b}{2 a}$ and $k = c - {b}^{2} / \left(4 a\right)$

From the given $y = 2 {x}^{2} - 4 x - 3$
$a = 2$ and $b = - 4$ and $c = - 3$

$h = \frac{- b}{2 a} = \frac{- \left(- 4\right)}{2 \left(2\right)} = 1$
$k = c - {b}^{2} / \left(4 a\right) = - 3 - {\left(- 4\right)}^{2} / \left(4 \left(2\right)\right) = - 5$

Axis of symmetry:

$x = h$

$\textcolor{b l u e}{x = 1}$

Since $a$ is positive, the function has a Minimum value and does not have a Maximum.

Minimum value $\textcolor{b l u e}{= k = - 5}$

The graph of $y = 2 {x}^{2} - 4 x - 3$

To draw the graph of $y = 2 {x}^{2} - 4 x - 3$, use the vertex $\left(h , k\right) = \left(1 , - 5\right)$ and the intercepts.

When $x = 0$,
$y = 2 {x}^{2} - 4 x - 3$
$y = 2 {\left(0\right)}^{2} - 4 \left(0\right) - 3 = - 3 \text{ }$means there is a point at $\left(0 , - 3\right)$

and when $y = 0$,
$y = 2 {x}^{2} - 4 x - 3$
$0 = 2 {x}^{2} - 4 x - 3$

$x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a} = \frac{- \left(- 4\right) \pm \sqrt{{\left(- 4\right)}^{2} - 4 \left(2\right) \left(- 3\right)}}{2 \left(2\right)}$

$x = \frac{+ 4 \pm \sqrt{16 + 24}}{4}$

$x = \frac{+ 4 \pm \sqrt{40}}{4}$

$x = \frac{+ 4 \pm 2 \sqrt{10}}{4}$

${x}_{1} = 1 + \frac{1}{2} \sqrt{10}$

${x}_{2} = 1 - \frac{1}{2} \sqrt{10}$

We have two points at $\left(1 + \frac{1}{2} \sqrt{10} , 0\right)$ and $\left(1 - \frac{1}{2} \sqrt{10} , 0\right)$

God bless...I hope the explanation is useful.