Answer
a) There are 2 points $(-2,16)$ and $(3,11)$ where the tangent is perpendicular to the given line.
b) There are 2 points $(0,20)$ and $(1,7)$ where the tangent is parallel to the given line.
Work Step by Step
$$y=2x^3-3x^2-12x+20$$
1) First, find the derivative of $y$: $$y'=2\times3x^2-3\times2x-12+0$$ $$y'=6x^2-6x-12$$
a) $(w):y=1-(x/24)$
The slope of $(w)$ is $-1/24$.
For the tangent to be perpendicular to $(w)$, the product of the slopes of the two lines must equal $-1$. In other words, $$y'\Big(-\frac{1}{24}\Big)=-1$$ $$y'=\frac{-1}{-\frac{1}{24}}=24$$ $$6x^2-6x-12=24$$ $$6x^2-6x-36=0$$ $$x^2-x-6=0$$ $$(x+2)(x-3)=0$$ $$x=-2\hspace{1cm}\text{or}\hspace{1cm}x=3$$
- For $x=-2$: $$y=2(-2)^3-3(-2)^2-12(-2)+20$$ $$y=-16-12+24+20=16$$
- For $x=3$: $$y=2(3)^3-3(3)^2-12(3)+20$$ $$y=54-27-36+20=11$$
So there are 2 points $(-2,16)$ and $(3,11)$ where the tangent is perpendicular to the given line.
b) $(r):y=\sqrt2-12x$
The slope of $(r)$ is $-12$.
For the tangent to be perpendicular to $(r)$, their slopes must be equal. In other words, $$y'=-12$$ $$6x^2-6x-12=-12$$ $$6x^2-6x=0$$ $$6x(x-1)=0$$ $$x=0\hspace{1cm}\text{or}\hspace{1cm}x=1$$
- For $x=0$: $$y=2(0)^3-3(0)^2-12(0)+20=20$$
- For $x=1$: $$y=2(1)^3-3(1)^2-12(1)+20$$ $$y=2-3-12+20=7$$
So there are 2 points $(0,20)$ and $(1,7)$ where the tangent is parallel to the given line.
