# How do you find the vertex and axis of symmetry, and then graph the parabola given by: y= -2x^2?

Oct 1, 2015

Vertex: $\left(0 , 0\right)$
Axis of symmetry: $x = 0$
(See below for graph)

#### Explanation:

$y = - 2 {x}^{2}$
and be re-written in explicit vertex form as
$y = \left(- 2\right) {\left(x - \textcolor{red}{0}\right)}^{2} + \textcolor{b l u e}{0}$
$\textcolor{w h i t e}{\text{XXX}}$with vertex at $\left(\textcolor{red}{0} , \textcolor{b l u e}{0}\right)$

Any parabola with the form:
$\textcolor{w h i t e}{\text{XXX}} y = a {x}^{2} + b x + c$
has a vertical axis of symmetry (through the vertex)
and opens upward if $a > 0$ or downward if $a < 0$
Therefore the axis of symmetry for $y = - 2 {x}^{2} \left(+ 0 x + 0\right)$
is $x = 0$
(and opens downward).

Note that
$\textcolor{w h i t e}{\text{XXX}}$x-intercept (value of $x$ when $y = 0$) is $0$
and
$\textcolor{w h i t e}{\text{XXX}}$y-intercept (value of $y$ when $x = 0$) is $0$
so
$\textcolor{w h i t e}{\text{XXX}}$the x and y intercepts do not give use any additional point to help us plot the graph.
Therefore it will be necessary to pick a few other values for $x$ and evaluate their corresponding $y$ values in order to plot the graph.

{: (x,,y), (0,,0), (-1,,-2), (+1,,-2), (-2,,-8), (+2,,-8) :}

By plotting these point-pairs on the Cartesian plane we should be able to sketch the graph for this function.

(Unfortunately, at this time, there seems to be a bug with Socratic.org that prevents posting a graph. I will try to check again later and add the graph if possible).