## In this exploration, we will be looking at the polar equation r = a + bcos(k) for various values of a, b, and. To begin, let’s look at a very basic form of the equation, that is, let a and b be equal to 1 and let k equal 5 (see the graph below).

## We see that the result is a flower-shaped graph with 5 petals, and we see that the y-intercepts are at -1 and +1. Notice that the y-intercepts are equal to ±a/b. Let’s see if this holds true for other values of a and b (but keeping a equal to b). Below are the graphs of our polar equation for the cases when a and b are equal to 2, 5, and 10, respectively (in each case, k is still equal to 5).

## Notice the difference in scale for each of these graphs; the shape remains the same as we change a and b, but the size of the flower increases as a and b increase. For each of the three cases shown above, we do still see that the y-intercepts are equal to ±a/b.

## Notice also from the original graph shown above (a=b=1 and k=5) that the number of petals on our flower is equal to the parameter k. Let’s see if this holds true for other values of k. Below are the graphs of our polar equation for the cases when k is equal to 1, 3, and 6 respectively (in each case, a and b are both equal to 1).

## We see here that the number of petals does appear to always be equal to the parameter k, but notice the case when k is equal to 6. The y-intercepts are no longer equal to ±a/b. This is actually true for any

evenvalues of a and b.## Let’s turn our attention now to the polar equation r = bcos(k) and see how it behaves in relation to our first equation. Below is the graph for the case when b is equal to 1 and k is equal to 3.

## Notice that there are still k leafs in our graph, but now instead of the y-intercept being equal to ±b, the x-intercept is equal to just +b. Let’s see if these relationships hold true for other values of k and b. Below you will see the graph of our equation for the cases when k is equal to 2, 4, and 5 respectively (in each case, the b value is 1).

## We see that when k is equal to both 2 and 4, the number of petals is equal to 2k. This is actually true for all even values of the parameter k. When k is odd, the number of petals is still equal to k, as with the first equation.

## Now let’s look at what happens when k is held constant and b is changed. Below you will see the graphs for the cases when b is equal to 2, 5, and 8 respectively (for each case, k is equal to 3).

## We see that the shape of the graph stays the same for each value of b, but the size increases (notice that the scale in each graph is different). An x-intercept in each case is equal to b, as in the first case. Refer back to the graphs for varying values of k and notice that when k is even, there are two x-intercepts, and they are at +b and –b.

## Lastly, let’s see what happens when cosine is replaced with sine for both of our equations (r = a + bcos(k) and r = bcos(k)).

## Below are the graphs for the equations r = 1 + 1sin(3) and r = 1sin(3).

## Notice that the overall shape and size of the graphs is the same as for the cases with cosine, but the graph has been rotated around the origin 90 degrees.

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