91Ó°ÊÓ

We have:

$y=x^2+6x+13$

First, calculate the $x-$intercept by putting $f\left(x\right)=0$:

$x^2+6x+13=0$

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

$x_{1, 2}=\frac{-bÂ\pm \sqrt{b^2-4ac}}{2a}$

For $a=1, b=6, c=13$, it follows:

$$\begin{gathered} x_{1, 2}=\frac{-6Â\pm \sqrt{6^2-4· 1· 13}}{2· 1} \\ {x}_{1, 2}=\frac{-6Â\pm  4i}{2· 1} \\ x=-3+2i, x=-3-2i \end{gathered}$$

There is no $x-$intercept: $x∉R$.

Now, calculate the $y-$intercept by putting $x=0:$

$$\begin{gathered} f\left(0\right)={\left(0\right)}^2+6\left(0\right)+13 \\ f\left(0\right)=13 \end{gathered}$$

$y-$intercept:$(0,13)$

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

The parabola params are:

$$\begin{gathered} a=1, b=6, c=13 \\ {x}_v=-\frac{b}{2a} \\ {x}_v=-\frac{6}{2· 1} \\ {x}_v=-3 \end{gathered}$$

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

$$\begin{gathered} y_v=4 \\ \mathrm{Therefore} \mathrm{the} \mathrm{parabola} \mathrm{vertex} \mathrm{is} \\ \left(-3, 4\right) \\ \mathrm{If} \mathrm{a}<0, \mathrm{then} \mathrm{the} \mathrm{vertex} \mathrm{is} \mathrm{a} \mathrm{maximum} \mathrm{value} \\ \mathrm{If} \mathrm{a}>0, \mathrm{then} \mathrm{the} \mathrm{vertex} \mathrm{is} \mathrm{a} \mathrm{minimum} \mathrm{value} \\ \mathrm{a}=1 \\ \mathrm{Minimum}\left(-3, 4\right) \end{gathered}$$

We have:

$f\left(x\right)=x^2+6x+13$

Identify the coefficients:

$a=1,b=6$

Substitute the coefficients into the expression:

$$\begin{gathered} x=-\frac{6}{2×1} \\ x=-3 \end{gathered}$$

Plot the y-intercepts on the graph then plot the vertex at $\left(-3,4\right)$and then draw the axis of symmetry: