Answer
The graph is shown below.
Range $[-6,\infty)$.
Work Step by Step
The given function is a quadratic function:
$f(x)=x^2+6x+3$
The standard form of the quadratic function is
$f(x)=ax^2+bx+c$
Compare both equations $a=1,b=6$ and $c=3$.
Step 1:- Parabola opens.
$a>0$, the parabola open upward.
Step 2:- Vertex.
$x-$coordinate of the vertex is $x=-\frac{b}{2a}=-\frac{(6)}{2(1)}=-3$.
Substitute the value of $x$ into the function.
$\Rightarrow f(-3)=(-3)^2+6(-3)+3$
Simplify.
$\Rightarrow f(-3)=9-18+3$
$\Rightarrow f(-3)=-6$
The vertex is $(-3,-6)$.
Step 3:- $x-$intercepts.
Replace $f(x)$ with $0$ into the given function.
$\Rightarrow 0=x^2+6x+3$
Use quadratic formula, we have $a=1,b=6$ and $c=3$.
$\Rightarrow x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$
Substitute all values.
$\Rightarrow x=\frac{-(6)\pm\sqrt{(6)^2-4(1)(3)}}{2(1)}$
Simplify.
$\Rightarrow x=\frac{-6\pm\sqrt{36-12}}{2}$
$\Rightarrow x=\frac{-6\pm\sqrt{24}}{2}$
$\Rightarrow x=\frac{-6\pm2\sqrt6}{2}$
Separate the fractions.
$\Rightarrow x=\frac{-6+2\sqrt6}{2}$ or $x= \frac{-6-2\sqrt6}{2}$
Simplify.
$\Rightarrow x=-3+\sqrt6$ or $ x=-3-\sqrt6$
The $x-$intercepts are $-3+\sqrt6$ and $-3-\sqrt6$. The parabola passes through $(-3+\sqrt6,0)$ and $(-3-\sqrt6,0)$.
Step 4:- $y-$intercept.
Replace $x$ with $0$ in the given function.
$\Rightarrow f(0)=(0)^2+6(0)+3$
Simplify.
$\Rightarrow f(0)=3$
The $y-$intercept is $3$. The parabola passes through $(0,3)$.
Step 5:- Graph.
Use the points:- vertex, $x-$intercepts and $y-$intercept to draw a parabola.
The axis of symmetry is $x=-3$.
$A=(-3,-6)$
$B=(-3+\sqrt6,0)$
$C=(-3-\sqrt6,0)$
$D=(0,3)$.
From the graph the range of the function is
$[-6,\infty)$.
