Given y=-2(x+4)^2-3, how do you find the vertex and axis of symmetry?

Jan 9, 2017

$\text{vertex " =(-4,-3)," axis of symmetry is } x = - 4$

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.

$y = - 2 {\left(x + 4\right)}^{2} - 3 \text{ is in this form}$

$\text{ Note } {\left(x + 4\right)}^{2} = {\left(x - \left(- 4\right)\right)}^{2}$

$\text{by comparison " h=-4" and } k = - 3$

$\Rightarrow \text{ vertex } = \left(- 4 , - 3\right)$

The axis of symmetry passes through the vertex.

Since the coefficient of ${\left(x + 4\right)}^{2} \text{ is negative, then the parabola opens down}$

$\Rightarrow \text{equation of vertex is } x = - 4$
graph{-2(x+4)^2-3 [-10, 10, -5, 5]}