Question

Determine if the graph is symmetric about the x-axis, the y-axis, or the origin.
r=4cos5 (theta)

Answer 1

Can you show how?

Answer 2

The equation is symmetric about the y-axis. This occurs when f(x) = f(-x) . Okay, that's the abstraction. For your particular function:

cos (theta) = cos(-theta)

What this actually means is that if you start at zero degrees and move, let's say, less than 90 clockwise or counter-clockwise. You will be respectively in either the 4th or the first quadrant where you have "x" as positive. Likewise, if you are moving between 90 and 180 degrees, you will be in the second and third quadrants where you have -x in both cases.

Consider the function y = x^2. Whether you put x or -x, you get the same positive "y". The cos function is the same way. Symmetric about the y-axis.

Answer 3

Here's a graph from which you can see this symmetry about the y-axis:

Answer 4

So basically if r=4cos (-5theta) its rhe same as the equation above?

Answer 5

A graph is said to be symmetric about the x-axis if whenever (a,b) is on the graph then so is (a,-b)

use the same logic for the y axis.

A graph is said to be symmetric about the y-axis if whenever (a,b) is on the graph then so is (-a,b)

Answer 6

yes, 4 cos(-5 theta) = 4 cos(+5 theta)

Answer 7

Oh, ok! Thank you!

Answer 8

uw! Good luck in all of your studies and thx for the recognition! @Catsarecool97 (even though I'm a dog-lover!)

Answer 9

Just a side note about that function:

r = 4cos (5 theta) .

I answered that with the function being non-polar. But if this function actually is polar, then there is no symmetry. Usually if a function is polar, the problem has to state that it is. Here, it's a little confusing because you are using "r" and "theta", which makes it seem like it might be polar, but the problem doesn't state that. So . . .

If polar: no symmetry
If not polar: symmetry about the y-axis.

Hope that helps.

Answer 10

Actually, if it is polar, it is still symmetric. It is a rose.

Answer 11

You want to look into this:
http://www.egr.msu.edu/~wandyezj/site_subpages/MTH133/MTH133_11.4_Graphing_in_polar.pdf

Answer 12

divide 2pi in 5 parts