Question

test for symmetry and graph the polar equation.
r = 6 + 6\cos\theta
a. is the polar equation symmetrical 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 polar equation symmetrical with respect to the line \theta=\frac{\pi}{2}?
a. yes.
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. 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}.
c. is the polar equation symmetrical with respect to the pole?
a. the polar equation failed the test for symmetry which means that the graph is not symmetric with respect to the pole.
b. the polar equation failed the test for symmetry which means that the graph may or may not be symmetric with respect to the pole.
c. yes.

Explanation:

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

Replace $\theta$ with $-\theta$.
Since $\cos(-\theta)=\cos\theta$, the equation $r = 6+6\cos(-\theta)=6 + 6\cos\theta$ remains the same. So, it is symmetric about the polar axis.

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

Replace $(r,\theta)$ with $(-r,-\theta)$.
The new equation is $-r=6 + 6\cos(-\theta)=6+6\cos\theta$, or $r=-6 - 6\cos\theta$, which is not the same as the original equation. Also, replacing $\theta$ with $\pi-\theta$ gives $r = 6+6\cos(\pi-\theta)=6-6\cos\theta
eq6 + 6\cos\theta$. So, it may or may not be symmetric about $\theta=\frac{\pi}{2}$.

Step3: Test for symmetry about the pole ($r = 0$)

Replace $r$ with $-r$. The new equation is $-r=6+6\cos\theta$, or $r=-6 - 6\cos\theta$, which is not the same as the original equation. So, it may or may not be symmetric about the pole.

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. B. The polar equation failed the test for symmetry which means that the graph may or may not be symmetric with respect to the pole.