Now consider the following polar equation:

Let's first consider the case where *k* is odd, as in the
equations

Looking at the graph, we see that *k* determines the number
of leaves, and each leaf is *b* units long. They are equally spaced
around the origin, and the end of one of the leaves is on the positive x-axis.

When *b* is negative, the graph is rotated /*k*
radians from the graph when *b* is positive:

Now let *k* be even, as in the equations

We see from the graph that the number of leaves is 2*k*,
and each leaf is *b* units long.

When *b* is negative, the graph is the same.

Conclusions about the graph of the polar equation

When *k* is odd, the rose has *k* leaves.

When *k* is even, the rose has 2*k* leaves.

The leaves are equally spaced about the origin.

The length of each leaf (the distance from the origin to
the point farthest from the origin) is | *b* |.

When *k* is odd and *b* is negative, the graph is
rotated /*k* radians
from the graph when *b* is positive.

When *k* is even and *b* is negative, the graph is the
same as the graph when *b* is positive.