# How do find the vertex and axis of symmetry, and intercepts for a quadratic equation y= -2x^2?

Aug 21, 2015

vertex at $\left(0 , 0\right)$
axis of symmetry: $x = 0$
y-intercept: $0$
x-intercept: $0$

#### Explanation:

General vertex form for a quadratic is
$\textcolor{w h i t e}{\text{XXXX}} y = m \left(x - a\right) + b$
$\textcolor{w h i t e}{\text{XXXXXXXX}}$with the vertex at $\left(a , b\right)$

$y = - 2 {x}^{2}$ can be written in vertex form as
$\textcolor{w h i t e}{\text{XXXX}} y = \left(- 2\right) {\left(x - 0\right)}^{2} + 0$
$\textcolor{w h i t e}{\text{XXXXXXX}}$with vertex at $\left(0 , 0\right)$

The equation is that of a standard parabola (opening downward)
so the axis of symmetry is a vertical line passing through the vertex
i.e. $x = 0$

The y-intercept is the value of $y$ when $x = 0$

The x-intercept(s) is/are the value of $x$ when $y = 0$ (in this case there is only one solution to $0 = - 2 {x}^{2}$)