Answer
See below:
Work Step by Step
To determine the graph, at first we will check the symmetry across the polar axis, the line $\theta =\frac{\pi }{2}$ and across the pole.
To find symmetry across the polar axis, we will replace $\theta $ with $-\theta $,
Then,
$\begin{align}
& r=1+\sin \left( -\theta \right) \\
& =1-\sin \theta
\end{align}$
So it is not symmetrical across the polar axis because $ r\ne 1+\sin \theta $.
Now, we will check across the line $\theta =\frac{\pi }{2}$; then we will replace $\left( r,\theta \right)=\left( -r,-\theta \right)$.
$\begin{align}
& -r=1+\sin \left( -\theta \right) \\
& -r=1-\sin \theta \\
& r=-1+\sin \theta
\end{align}$
So it is not symmetrical across the line $\theta =\frac{\pi }{2}$ because $ r\ne 1+\sin \theta $.
Now we will check symmetry across the pole by replacing $ r\,\,\text{with}\,\,-r $,
$\begin{align}
& -r=1+\sin \theta \\
& r=-1-\sin \theta \\
\end{align}$
So it is not symmetrical across the pole because $ r\ne 1+\sin \theta $.
So from the above assumption, it has not satisfied any kind of symmetry.
Now we will plot the graph with the range taken as the period of $\sin \theta $, $\left[ 0,2\pi \right]$; thus the value of $\theta $ will vary from $0\,\,\text{to}\,\,2\pi $.
Put $\theta =0{}^\circ $ in $ r=1+\sin \theta $,
Then,
$\begin{align}
& r=1+\sin 0{}^\circ \\
& =1+0 \\
& =1
\end{align}$
So at $\theta =0$, $ r=1$.
Similarly, we put in values between $0\,\,\text{to}\,\,2\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 $ $\frac{7\pi }{6}$ $\frac{4\pi }{3}$ $\frac{3\pi }{2}$ $\frac{5\pi }{3}$ $\frac{11\pi }{6}$ $2\pi $
$ r $ 1 1.5 1.87 2 1.87 1.5 1 0.5 0.13 0 0.13 0.5 1
