## Precalculus (6th Edition) Blitzer 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