Answer
Symmetric about the x-axis.
No symmetry about the y-axis.
No symmetry about the origin.
Work Step by Step
Testing for symmetry across the x-axis
(replace $\theta $ with $-\theta$ and see we arrive at an equivalent equation):
$ r=1+\cos(-\theta)\qquad$ ... cosine is an even function,
$r=1+\cos\theta\qquad $... we have symmetry about the x-axis.
Testing for symmetry across the y-axis:
1. replace $\theta $ with $-\theta$ and $r$ with $-r;$ see we arrive at an equivalent equation)
$-r=1+\cos(-\theta)\qquad$ ... cosine is an even function,
$-r=1+\cos\theta\qquad$ ... not equivalent.
2. replace $\theta $ with $\pi-\theta ;$ see that we arrive at an equivalent equation)
$ r=1+\cos(\pi-\theta)\qquad$ ... apply the trigonometric identity,
$ r=1-\cos\theta\qquad$ ... not equivalent.
Neither $(r, \pi-\theta)$ nor $(-r, -\theta)$ lie on the curve when $(r,\theta)$ lies on the curve.
No symmetry about the y-axis.
No symmetry about the origin, since it is not symmetric about both axes.
To graph, select a few values for $\theta$ and calculate $r.$
Use the symmetry.
See the resulting image.
