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

Oct 9, 2017

See explanation...

#### Explanation:

The given expression:

$0 = {x}^{2} + 6 x + 9$

is an equation, not a function.

We can express the related function as:

$f \left(x\right) = {x}^{2} + 6 x + 9 = {\left(x + 3\right)}^{2}$

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

Note that for any real number $t$, we have ${t}^{2} \ge 0$, with equality if and only if $t = 0$.

Hence:

${\left(x + 3\right)}^{2} \ge 0$

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 $\left(- 3 , 0\right)$

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

$f \left(0\right) = {0}^{2} + 6 \left(0\right) + 9 = 9$

That is: $\left(0 , 9\right)$

We could evaluate $f \left(x\right)$ 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]}

Oct 9, 2017

$y = {\left(x + 3\right)}^{2}$
graph{(x+3)^2 [-10, 10, -5, 5]}

#### Explanation:

x^2+6x+9=x^2+(2*(3x))+3^2)
This is in the form ${x}^{2} + \left(2 x y\right) + {y}^{2}$ where $x = x$ and $y = 3$
We know ${x}^{2} + \left(2 x y\right) + {y}^{2} = {\left(x + y\right)}^{2}$
$\therefore {x}^{2} + \left(2 \cdot \left(3 x\right)\right) + {3}^{2} = {\left(x + 3\right)}^{2}$
I.e. $y = {\left(x + 3\right)}^{2}$