Answer
See below:
Work Step by Step
To determine the graph, we will check the symmetry across the polar axis, the line $\theta =\frac{\pi }{2}$ and across the pole.
To find the symmetry across the polar axis, we will replace $\theta $ with $-\theta $.
Then,
$\begin{align}
& r=1+3\cos \left( -\theta \right) \\
& =1+3\cos \theta
\end{align}$
So, it is symmetrical across the polar axis because $ r=1+3\cos \theta $.
Now we will check across the line $\theta =\frac{\pi }{2}$ then replace $\left( r,\theta \right)=\left( -r,-\theta \right)$,
$\begin{align}
& -r=1+3\cos \left( -\theta \right) \\
& -r=1+3\cos \theta \\
& r=-1-3\cos \theta
\end{align}$
So, it is not symmetrical across the line $\theta =\frac{\pi }{2}$ because $ r\ne 1+3\cos \theta $.
Now we will check the symmetry across the pole by replacing $ r\,\,\text{with}\,\,-r $,
$\begin{align}
& -r=1+3\cos \theta \\
& =-1-3\cos \theta
\end{align}$
So, it is not symmetrical across the pole because $ r\ne 1+3\cos \theta $.
So, from the above, we have symmetry across the polar axis but not across the line $\theta =\frac{\pi }{2}$ and not across the pole.
Now to plot the graph in the range taken between $\left[ 0,\pi \right]$, use the symmetry across the polar axis.
Put $\theta =0{}^\circ $ in $ r=1+3\cos \theta $,
Then,
$\begin{align}
& r=1+3\cos 0{}^\circ \\
& =1+3 \\
& =4
\end{align}$
So at $\theta =0$, $ r=4$.
Similarly, put the values between $0\,\,\text{to}\,\,\pi $ and find the values of r seen below,
$\theta $ 0 $\frac{\pi }{6}$ $\frac{\pi }{3}$ $\frac{\pi }{2}$ $\frac{2\pi }{3}$ $\frac{5\pi }{6}$ $\pi $
$ r $ 4 3.6 2.5 1 -0.5 -1.6 -2
