Answer:
Both equation are symmetric about the x-axis.
Step-by-step explanation:
If (r,θ) can be replaced by (r,-θ),then the graph is symmetric about the x-axis.
If (r,θ) can be replaced by (-r,-θ),then the graph is symmetric about the y-axis.
If (r,θ) can be replaced by (-r,θ),then the graph is symmetric about the origin.
The given equation is
[tex]r=9\cos(5\theta)[/tex]
Replace the value of (r,θ) by (r,-θ).
[tex]r=9\cos(5(-\theta))=9\cos(5\theta)=r[/tex]
It is symmetric about the x-axis.
Replace the value of (r,θ) by (-r,-θ).
[tex]-r=9\cos(5(-\theta))=9\cos(5\theta)=r\neq -r[/tex]
It is not symmetric about the y-axis.
Replace the value of (r,θ) by (-r,θ).
[tex]-r=9\cos(5(\theta))=r\neq -r[/tex]
It is not symmetric about the origin.
Therefore the first equation is symmetric about the x-axis.
The given equation is
[tex]r=2\cos(\theta)[/tex]
Replace the value of (r,θ) by (r,-θ).
[tex]r=2\cos(-\theta)=2\cos(\theta)=r[/tex]
It is symmetric about the x-axis.
Replace the value of (r,θ) by (-r,-θ).
[tex]-r=2\cos(-\theta)=2\cos(\theta)=r\neq -r[/tex]
It is not symmetric about the y-axis.
Replace the value of (r,θ) by (-r,θ).
[tex]-r=2\cos(\theta)=r\neq -r[/tex]
It is not symmetric about the origin.
Therefore the second equation is symmetric about the x-axis.
