Answer
See image:
Work Step by Step
$\begin{array}{llll}
\text{Function:} & f(x)=x^{2}+3x+2 & a =1, b=3, c=2 & \\
\text{coefficient a} & \text{positive, opens up} & & \\
\text{Vertex} & x_{V}=-b/(2a) & y_{V}=f(-b/(2a) & \\
& x_{V}=-3/2 & y_{V}=-1/4 & \\
& \bf V(-3/2, -1/4) & & \\\\
\text{Line of symmetry} & x=-3/2 & & \\
\text{y intercept} & c=2, & \bf(0,2) & \\
\text{point symmetric to } & & & \\
\text{the y-intercept} & x=2x_{V}=-3 & \bf(-3,2)\\ & \\
\text{Zeros:} & x^{2}+3x+2=0 & & \\
& (x+1)(x+2)=0 & & \\
& x=-1,x=-2 & & \\
& \bf(-2,0), (-1,0) & &
\end{array}$
Using the above information, plot the five points and join with a smooth curve (parabola).