## University Calculus: Early Transcendentals (3rd Edition)

There are 2 points $(-1,27)$ and $(2,0)$ for which the tangents there are parallel to the $x-$axis.
$$y=2x^3-3x^2-12x+20$$ 1) First, find the derivative of $y$: $$y'=2\times3x^2-3\times2x-12+0$$ $$y'=6x^2-6x-12=6(x^2-x-2)$$ 2) All the tangents being parallel to the $x-$axis, or in other words, horizontal tangents, have zero slopes. This means we will see if there are any values of $x$ for which $y'=0$ or not; these values of $x$ represent points where the tangents are horizontal. $$y'=0$$ $$6(x^2-x-2)=0$$ $$x^2-x-2=0$$ $$(x+1)(x-2)=0$$ $$x=-1\hspace{1cm}\text{or}\hspace{1cm}x=2$$ - For $x=-1$: $$y=2(-1)^3-3(-1)^2-12(-1)+20$$ $$=2(-1)-3(1)+12+20$$ $$=-2-3+32=27$$ - For $x=2$: $$y=2(2)^3-3(2)^2-12(2)+20$$ $$=2^4-3(4)-24+20$$ $$=16-12-4=0$$ Therefore, there are 2 points $(-1,27)$ and $(2,0)$ for which the tangents there are parallel to the $x-$axis.