# Find specific tangent lines of $y=2x^3-3x^2-12x+20$

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 $$y=\sqrt{2}-12x$$.

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 $$y'=-12$$.