# How do you graph the quadratic function and identify the vertex and axis of symmetry for y=-(x-2)^2-1?

Jan 24, 2017

$\left(2 , - 1\right) , x = 2$

#### Explanation:

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

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

$y = - {\left(x - 2\right)}^{2} - 1 \text{ is in this form}$

and by comparison $h = 2 \text{ and } k = - 1$

$\Rightarrow \text{vertex } = \left(2 , - 1\right)$

Since the $\textcolor{b l u e}{\text{value of a is negative}}$

$\text{That is } \textcolor{red}{-} {\left(x - 2\right)}^{2}$

Then the parabola opens down vertically color(red)(nnn

The axis of symmetry goes through the vertex and therefore has equation $\textcolor{m a \ge n t a}{\text{x=2}}$

$\textcolor{b l u e}{\text{Intercepts}}$

$x = 0 \to y = - {\left(0 - 2\right)}^{2} - 1 = - 4 - 1 = - 5$

$\Rightarrow y = - 5 \leftarrow \textcolor{red}{\text{y-intercept}}$

$y = 0 \to - {\left(x - 2\right)}^{2} - 1 = 0 \to {\left(x - 2\right)}^{2} = - 1$

This has no real solutions hence there are no x-intercepts.

Knowing the ' shape' of the parabola, the vertex and the y-intercept enables the graph to be sketched.
graph{-(x-2)^2-1 [-10, 10, -5, 5]}