y =y/x^2 + ( x - 3 ) ( y + 1 )y=yx2+(x−3)(y+1). Reorganizing,
y(x^3 - 3 x^2 + 1 ) = - x^2 ( x - 3 )y(x3−3x2+1)=−x2(x−3)
Besides y = - 1y=−1, the asymptotes are given by{ x = root
of x^3 - 3x^2 + 1 = 0x3−3x2+1=0 }.
See graph for the roots.
graph{y -(x^3-3x^2+1)=0}
Using ( x, y ) = r ( cos theta, sin theta )(x,y)=r(cosθ,sinθ), this converts to
cos theta ( r^3 sin^3theta - 3 r^2 sin^2theta + 1 )cosθ(r3sin3θ−3r2sin2θ+1)
r sin theta - 3 )# Reorganizing,
r^3 sin theta cos^3thetar3sinθcos3θ
- r^2 cos^2theta ( 3 sin theta - cos theta ) - 3 r cos^2 theta−r2cos2θ(3sinθ−cosθ)−3rcos2θ
+ sin theta = 0+sinθ=0
The period of r = f ( sin thetar=f(sinθ or cos thetacosθ ) is 2pi2π.
Easily, on theta = 0, r = 0, 3.θ=0,r=0,3.
See graph, for the Cartesian frame.
graph{y(x^3 - 3 x^2 + 1 ) + x^2 ( x - 3 )=0}
Compare with the first graph roots, giving asymptotes x = root.
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