Choosing the graph of a limacon

Plot a few key angles for $r=3\cos \theta +2$: - $\theta =0$: $r=3+2=5$, the point 5 units out on the polar axis. - $\theta =\frac{\pi }{2}$: $r=0+2=2$, the point 2 units straight up. - $\theta =\pi$: $r=-3+2=-1$. - $\theta =\frac{3\pi }{2}$: $r=2$, the point 2 units straight down.

The negative value is what shapes the graph. A point with $r=-1$ at angle $\pi$ is plotted 1 unit in the opposite direction, on the positive polar axis at distance 1. $r$ is negative whenever $3\cos \theta +2<0$, that is $\cos \theta <-\frac{2}{3}$, an interval of angles around $\pi$. During that interval the curve passes through the origin, traces a small loop on the right that reaches out to 1, and comes back through the origin.

So the graph is a limaçon that reaches 5 on the polar axis, crosses the vertical axis at 2 above and below the pole, and has an inner loop reaching 1 on the polar axis. Choice D shows exactly that.

Each choice

• A. This curve reaches only 4 on the polar axis and crosses the vertical axis at 1, with no inner loop. That is $r=3\cos \theta +1$ drawn only where $r\ge 0$: the constant is off by one and the negative values were left out.
• B. An inner loop is right, but the outer curve reaches only 4 and the loop reaches 2. Those are the values of $r=3\cos \theta +1$ at $\theta =0$ and $\theta =\pi$. The given equation adds 2, which gives 5 and $-1$.
• C. The size is right, with 5 on the polar axis and 2 on the vertical axis, but there is no inner loop. This is what you get by ignoring the angles where $r$ is negative, such as $\theta =\pi$ where $r=-1$.
• D (correct). The only graph that matches: it reaches 5 at $\theta =0$, crosses the vertical axis at 2, and has an inner loop out to 1 from the angles where $r<0$.