Question

How do you graph the function `0 = x^2+6x+9` ?

Answer 1

See explanation...

Explanation:

The given expression:

`0 = x^2+6x+9`

is an equation, not a function.

We can express the related function as:

`f(x) = x^2+6x+9 = (x+3)^2`

In which case the given equation represents the zeros of the function, i.e. the intersections of `f(x)` with the `x` axis.

Note that for any real number `t`, we have `t^2>=0`, with equality if and only if `t=0`.

Hence:

`(x+3)^2 >= 0`

with equality if and only if `x+3 = 0`, i.e. `x = -3`

So what we have here is a parabola with vertex on the `x` axis at the point `(-3, 0)`

To find the intersection with the `y` axis, set `x=0` to find:

`f(0) = 0^2+6(0)+9 = 9`

That is: `(0, 9)`

We could evaluate `f(x)` for a few different values of `x` to find some more points through which the parabola passes, but note that the leading coefficient is `1`. So this parabola is essentially the same as `y=x^2`, but shifted `3` units to the left...

graph{x^2+6x+9 [-8, 3, -1.1, 10.2]}

Answer 2

`y=(x+3)^2`
graph{(x+3)^2 [-10, 10, -5, 5]}

Explanation:

`x^2+6x+9=x^2+(2*(3x))+3^2)`
This is in the form `x^2+(2xy)+y^2` where `x=x` and `y=3`
We know `x^2+(2xy)+y^2=(x+y)^2`
`:. x^2+(2*(3x))+3^2=(x+3)^2`
I.e. `y=(x+3)^2`