Answer
a. $(3,11)$ and $(-2,16)$
b. $(0,20)$ and $(1.7)$
Work Step by Step
a. Step 1.Given $y=2x^3-3x^2-12x+20$, we have $y'=6x^2-6x-12$ which gives the slope of tangent lines to the curve.
Step 2. Given the line equation $y=1-x/24$, we can find its slope as $m=-1/24$ and a line normal to this line will have a slope of $n=-1/m=24$
Step 3. Let $y'=n$, we have $6x^2-6x-12=24$ or $x^2-x-2=4$ which gives $x^2-x-6=0$ or $(x-3)(x+2)=0$, thus $x_1=3, x_2=-2$
Step 4. The corresponding y-values are $y_1=2(3)^3-3(3)^2-12(3)+20=11$ and $y_2=2(-2)^3-3(-2)^2-12(-2)+20=16$
Step 5. The points on the curve where the tangent is perpendicular to the line are $(3,11)$ and $(-2.16)$
b. Step 1. Given the line equation $y=\sqrt 2-12x$, we can find its slope as $m=-12$ and a line parallel to this line will have the same slope of $n=m=-12$
Step 2. Let $y'=n$, we have $6x^2-6x-12=-12$ or $x^2-x=0$ which gives $x_1=0, x_2=1$
Step 3. The corresponding y-values are $y_1=2(0)^3-3(0)^2-12(0)+20=20$ and $y_2=2(1)^3-3(1)^2-12(1)+20=7$
Step 5. The points on the curve where the tangent is parallel to the line are $(0,20)$ and $(1.7)$
