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]}