Answer
a. $y=-\frac{1}{8}x+\frac{5}{4}$
b. $-4$, $(0, 1)$
c. $y=8x-15$, $y=8x+17$
Work Step by Step
a. Step 1. Given the function equation $y=x^3-4x+1$ and the point $(2,1)$, we can check if the point is on the curve and find the slope of the tangent line as the derivative of the function at this point $m=y'|_{x=2}=(3x^2-4)|_{x=2}=3(2^2)-4=8$
Step 2. The slope of the line normal to the tangent line is the negative reciprocal of the original slope; we have $m'=-\frac{1}{m}=-\frac{1}{8}$
Step 3. Knowing the slope and the point, we can write the line equation as $y-1=-\frac{1}{8}(x-2)$ which gives $y=-\frac{1}{8}x+\frac{5}{4}$
b. From part-a, the function for the slope is $y'=3x^2-4$. We can find the minimum of this parabola at $x=0, y'=-4$; thus the smallest slope is $m_1=-4$ at point $(0, 1)$
c. Step 1. Let $m=8$, we have $3x^2-4=8$ and $x^2=4$ which gives $x=\pm2$
Step 2. For $x=2$, we can find the point on the curve $y=2^3-4(2)+1=1$, which is $(2,1)$. The slope of the tangent line is $m_2=3(2^2)-4=8$; thus the line equation is $y-1=8(x-2)$ or $y=8x-15$
Step 3. For $x=-2$, we can find the point on the curve $y=(-2)^3-4(-2)+1=1$, which is $(-2,1)$. The slope of the tangent line is $m_2=3(-2)^2-4=8$; thus the line equation is $y-1=8(x+2)$ or $y=8x+17$
