How do you graph `y=x^2+6x+9`?

The first task here is to find the roots of the equation, that is where the graph intercepts the `x` axis. To do this, set `y=0` and then factorise:

`x^2+6x+9=0`
`(x+3)^2=0 -> x=-3`

In this case as `x=-3` twice then the graph will "graze" x-axis at `x=-3`, (intercept it only once). Note, if there were 2 distinct roots then the graph would cut straight the through the `x` axis at the 2 points and if there were no real roots then the graph would not intercept the `x` axis at all.

Find where the graph intercepts the `y` axis by setting `x=0`

`-> y = (0)^2+6(0)+9=9`

So our `y` intercept is at `(0,9)`

We also have to find the turning point. This can be done in a number of ways, finding the midway point between the roots, setting the derivative equal to 0 or compete the square.

In this case we know that since `x=-3` is the only root then that must also be the turning point.

As it is a quadratic where the `x^2` coefficient is positive then we know the graph will be a parabola with a minimum.

We now have all the information we need to graph the function. Simply mark on your points and sketch:

graph{x^2+6x+9 [-10, 10, -2, 10]}