How do you graph y=x^2+6x+9y=x2+6x+9?

1 Answer
Feb 7, 2016

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

Explanation:

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

x^2+6x+9=0x2+6x+9=0
(x+3)^2=0 -> x=-3(x+3)2=0x=3

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

Find where the graph intercepts the yy axis by setting x=0x=0

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

So our yy intercept is at (0,9)(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=-3x=3 is the only root then that must also be the turning point.

As it is a quadratic where the x^2x2 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]}