91Ó°ÊÓ

We have:

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

x-intercept(s): It is the point where $y=f\left(x\right)=0$.

Substitute $f\left(x\right)=0$, it follows:

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

For a quadratic equation of the form $a{x}^2+bx+c=0$, the solutions are:

$$\begin{gathered} x_{1, 2}=\frac{-bÂ\pm \sqrt{b^2-4ac}}{2a} \\ \mathrm{For} a=1, b=6, c=9 \\ {x}_{1, 2}=\frac{-6Â\pm \sqrt{6^2-4· 1· 9}}{2· 1} \\ {x}_{1, 2}=\frac{-6Â\pm \sqrt{0}}{2· 1} \\ x=-3 \end{gathered}$$

y-intercept(s): It is the point where $y=0$, it follows:

$$\begin{gathered} y={\left(0\right)}^2+6\left(0\right)+9 \\ y=9 \end{gathered}$$

Therefore, the intercepts are:

$\mathrm{X} \mathrm{Intercepts}: \left(-3, 0\right), \mathrm{Y} \mathrm{Intercepts}: \left(0, 9\right)$

$\mathrm{The} \mathrm{vertex} \mathrm{of} \mathrm{an} \mathrm{up}-\mathrm{down} \mathrm{facing} \mathrm{parabola} \mathrm{of} \mathrm{the} \mathrm{form} y=a{x}^2+bx+c \mathrm{is} x_v=-\frac{b}{2a}$

The parabola params are:

$$\begin{gathered} a=1, b=6, c=9 \\ {x}_v=-\frac{b}{2a} \\ {x}_v=-\frac{6}{2· 1} \\ {x}_v=-3 \end{gathered}$$

Plug in $x_v=-3$to find the $y_v$:

$y_v=0$

Therefore, the parabola vertex is$\left(-3, 0\right)$.

The axis of symmetry of a parabola is a vertical line that divides the parabola into two congruent halves.

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

So, the line $x=-3$is the axis of symmetry.

Use intercepts, vertex, and the equation of axis of symmetry to draw the graph: