Question

How do you graph `r=8+6costheta`?

Answer 1

Using Socratic graphic utility, the graph of the limacon could be inserted, for the cartesian form `x^2+y^2=8sqrt(x^2+y^2)+6x`.

Explanation:

`r = 8 + 6 cos theta >=2.`

As`r(theta) = r(-theta)`, the graph is symmetrical about `theta = 0`.

Range of r is from 2 to 14. 1-period ( `theta in [-pi. pi]`) is inserted.

graph{x^2+y^2-6x-8sqrt(x^2+y^2)=0x^2 [-24, 24, -12, 12]}