Question

test for symmetry and then graph the polar equation.
r = 1 + 4\cos\theta
a. is the graph of the polar equation symmetric with respect to the polar axis?
a. the polar equation failed the test for symmetry which means that the graph is not symmetric with respect to the polar axis.
b. yes.
c. the polar equation failed the test for symmetry which means that the graph may or may not be symmetric with respect to the polar axis.
b. is the graph of the polar equation symmetric with respect to the line \theta=\frac{\pi}{2}?
a. the polar equation failed the test for symmetry which means that the graph is not symmetric with respect to the line \theta=\frac{\pi}{2}.
b. the polar equation failed the test for symmetry which means that the graph may or may not be symmetric with respect to the line \theta=\frac{\pi}{2}.
c. yes.
c. is the graph of the polar equation symmetric with respect to the pole?
a. the polar equation failed the test for symmetry which means that the graph may or may not be symmetric with respect to the pole.
b. the polar equation failed the test for symmetry which means that the graph is not symmetric with respect to the pole.
c. yes.

Explanation:

Step1: Test for symmetry about the polar axis ($\theta = 0$)

Replace $\theta$ with $-\theta$ in the equation $r = 1+4\cos\theta$.
Since $\cos(-\theta)=\cos\theta$, we get $r = 1 + 4\cos(-\theta)=1 + 4\cos\theta$.
The equation remains the same, so the graph is symmetric about the polar axis.

Step2: Test for symmetry about the line $\theta=\frac{\pi}{2}$

Replace $(r,\theta)$ with $(-r,-\theta)$ in the equation $r = 1+4\cos\theta$.
We get $-r=1 + 4\cos(-\theta)=1 + 4\cos\theta$, or $r=-1 - 4\cos\theta$, which is not the same as the original equation.
Also, replacing $r$ with $-r$ gives $-r = 1+4\cos\theta$ (not same as original) and replacing $\theta$ with $\pi-\theta$ gives $r = 1+4\cos(\pi - \theta)=1-4\cos\theta$ (not same as original). So, the test for symmetry about the line $\theta=\frac{\pi}{2}$ fails. But a failed test does not mean the graph is not symmetric.

Step3: Test for symmetry about the pole

Replace $r$ with $-r$ in the equation $r = 1+4\cos\theta$.
We get $-r=1 + 4\cos\theta$, or $r=-1 - 4\cos\theta$, which is not the same as the original equation.
Also, replacing $\theta$ with $\theta+\pi$ gives $r = 1+4\cos(\theta+\pi)=1-4\cos\theta$ (not same as original). So, the test for symmetry about the pole fails. But a failed test does not mean the graph is not symmetric.

Answer:

a. B. Yes.
b. B. The polar equation failed the test for symmetry which means that the graph may or may not be symmetric with respect to the line $\theta=\frac{\pi}{2}$.
c. A. The polar equation failed the test for symmetry which means that the graph may or may not be symmetric with respect to the pole.