0
$\begingroup$

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

$\endgroup$
1
$\begingroup$

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 $y'=-12$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.