Answer
a. horizontal $y=-4$. normal $x=1$, horizontal $y=0$. normal $x=-1$
b. slope $-3$, point $(0,-2)$, normal $y=\frac{1}{3}x-2$
Work Step by Step
a. Step 1. Given the function $y=x^3-3x-2$, the derivative of the function can be find as $y'=3x^2-3$
Step 2. A horizontal tangent will have a slope $m=0$. Let $y'=0$, we have $3x^2-3=0$ and $x=\pm1$
Step 3. For $x=1$, $y=1^3-3(1)-2=-4$, which gives the point $(1,-4)$ and a line equation of $y=-4$.
The line normal to this tangent is given by $x=1$.
Step 4. For $x=-1$, $y=(-1)^3-3(-1)-2=0$, which gives point $(-1,0)$ and a line equation of $y=0$.
The line normal to this tangent is given by $x=-1$
b. Step 1. The smallest slope on the curve can be found when $y'$ has its minimum which happens at $x=0, y'_1=-3$
Step 2. The point on the curve for this slope is given as $x=0, y=-2$ or point $(0,-2)$
Step 3. The line that is perpendicular to the curve’s tangent at this point has a slope of $m'=-(1/(-3))=1/3$ and its equation can be written as $y+2=\frac{1}{3}(x-0)$ or $y=\frac{1}{3}x-2$
