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 $y'=-12$.


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