Question

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

Answer 1

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

Explanation:

Lets start with the known information.

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

`(-2,-3)` 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,

`(x+2)^2-3 = 0`
`therefore (x+2)^2=3`
`therefore x+2= +-sqrt(3)`
`therefore x =-2+-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 (x+2)^2-3 = x^2+4x+4-3 = x^2+4x+1`

This `implies` that the y intercept `=(0,1)`
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]}