Answer
a) 8
b) $y=-1x+1$
$y=\frac{1}{3}x+2$
c) $x_1=1.58$
$x_2=0.423$
d) See image, try changing other values
Work Step by Step
a) There are 8 horizontal asymptotes
b) $y(y^2-1)(y-2)=x(x-1)(x-2)$
$y^4-2y^3-y^2+2y=x^3-3x^2+2x$
$\frac{dy}{dx}(4y^3-6y^2-2y+2)=3x^2-6x+2$
$\frac{dy}{dx}=\frac{3x^2-6x+2}{4y^3-6y^2-2y+2}$
At point $(0,1)$, $\frac{dy}{dx}=\frac{2}{-2}=-1\hspace{1cm}y=-1x+1$
At point $(0,2)$, $\frac{dy}{dx}=\frac{2}{6}=\frac{1}{3}\hspace{1.5cm}y=\frac{1}{3}x+2$
c) $3x^2-6x+2=0$
$x=\frac{6\pm\sqrt{36-24}}{6}=\frac{3\pm\sqrt{3}}{3}$
$x_1=1.58$
$x_2=0.423$
d) See image
