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

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Answer 1 · 2017-01-05T15:32:44

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

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

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]}

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