## Investigate r = a + b cos(kθ) for various values of a, b, and k.

## Lets start by setting a and b equal to 1 and observing how different values of k affect the graph of r = a + b cos(kθ) .

## r = 1 + 1 cos(2θ)

## r = 1 + 1 cos(3θ)

## r = 1 + 1 cos(4θ)

## r = 1 + 1 cos(5θ)

Our parameter k seems to represent the number of pedals. ± a/b appears to represent the y-intercepts of the graphs with an odd number of petals.

## Lets see if this holds for other values of a and b when k =5.

## a = b = 2

## r = 2 + 2 cos(5θ)

## a = b = 5

## r = 5 + 5 cos(5θ)

## a = b = 10

## r = 10 + 10 cos(5θ)

As a/b increase the shape of the graph stays the same, but this size of rose increases.

## We can conclude:

## Parameter k represents the number of petals in the rose.

## When k is odd and a=b, ± a/b represent the y-intercepts.

## When k is even, ± a/b does not represent the y-intercepts.

## Next, lets looks at r = b cos(kθ) for various values of b and k.

## Let b=1 and k=3

## r = 1 cos(3θ)

Parameter k still seems to represent the number of petals, but +b appears to represent the x-intercept instead of the y-intercept.

## Lets look at what happens when we change b, but hold k constant.

## The video below illustrates -5 ≤ b ≤ 5.

## Lets look at what happens when we change k and hold b constant.

## r = 1 cos(2θ)

## r = 1 cos(5θ)

## r = 1 cos(6θ)

Notice that when parameter k is even, there are 2k number of petals. Additionally, x-intercepts appear at –b and +b when k is even.

## We can conclude:

## When parameter k is odd, k represents the number of petals in the rose.

## +b represents the x-intercept.

## When parameter k is even, 2k represents the number of petals in the rose.

## – b and +b represent the x-intercepts.

## What do you think will happen if cos(kθ) is replaced with sin(kθ) in both equations.

## r = 1 + 1 sin(3θ)

## r =1 sin(3θ)

The overall shape and size are the same, but it appears that the graph has been rotated 90° to the right.## Click here to return to Lacy's homepage.