Answer
a) See image
b) 8 horizontal tangents
$x=0; x=0.5; x=1$
Work Step by Step
a) See image
b) $2 y^3+y^2-y^5=x^4-2 x^3+x^2$
$\frac{dy}{dx}(6y^2+2y-5y^4)=4x^3-6x^2+2x$
$\frac{dy}{dx}=\frac{4x^3-6x^2+2x}{6y^2+2y-5y^4}$
$4x^3-6x^2+2x=0\hspace{0.2cm}\to\hspace{0.2cm} x(2x^2-3x+1)=0$
$x(2x-1)(x-1)=0$
$x=0; x=0.5; x=1$
8 horizontal tangents
