# How do you graph and label the vertex and axis of symmetry of y=(x+2)^2-3?

Jan 4, 2018

Using the information given to you and a bit of expanding.

#### Explanation:

From this we can identify the the vertex of the graph which as we know would be

$\left(- 2 , - 3\right)$ The X value is negative as the bracket is equal to 0.

We can also work out the roots by solving for x with the given completed square. There set y = to 0 and solve,

${\left(x + 2\right)}^{2} - 3 = 0$
$\therefore {\left(x + 2\right)}^{2} = 3$
$\therefore x + 2 = \pm \sqrt{3}$
$\therefore x = - 2 \pm \sqrt{3}$

Now we have our roots of the graph.

Now finally its time for the y-intercept, this can easily be obtained by expanding out the given completed square.

$\therefore {\left(x + 2\right)}^{2} - 3 = {x}^{2} + 4 x + 4 - 3 = {x}^{2} + 4 x + 1$

This $\implies$ that the y intercept $= \left(0 , 1\right)$
Now all we need to do is plot the key points and sketch the graph. graph{y=(x+2)^2-3 [-10, 10, -5, 5]}