I am not sure where to begin with this problem. I need to find various specific tangent lines for a given function. I think I can solve #1 by setting $f'(x)=0$, and finding extrema, but I am not sure about that.
How can I solve this problem?
Find the points on the curve $y=2x^3-3x^2-12x+20$ where the tangent line is:
(1) parallel to the $x$-axis.
(2) perpendicular to $y=1-\frac{x}{24}.$
(3) parallel to $y=\sqrt{2}-12x$.
Best Answer
The derivative is $y'=6x^2-6x-12$. The points where the tangent line is parallel to the $x$-axis correspond to when $y'=0$. This is a quadratic equation, so you will be able to find at most two such points.
The points where the tangent line is perpendicular to $1-\frac{x}{24}$ are when $y'=24$ (perpendicular lines have negative reciprocal slopes). Once again, the equation $y'=24$ is a quadratic with at most two solutions.
The third case is similar, but now you want $y'=-12$.