How do you find the points where the graph of y^2+2y=4x^3−16x−1 is the tangent line vertical?

1 Answer
Apr 13, 2018

The coordinates of the points that have vertical tangents are at (-2, -1), (0, -1), and (2, -1).

Explanation:

If

y^2+2y=4x^3-16x-1

Then by implicit differentiation we have

2y(dy)/(dx)+2(dy)/(dx)=12x^2-16.

Now we can solve for (dy)/(dx).

(dy)/(dx)=(12x^2-16)/(2y+2)

The tangent line will be vertical when (dy)/(dx) becomes infinite. This will happen when

2y+2=0.

So the tangent line will be vertical when y=-1.

Now we need to calculate the x-coordinate(s) when y=-1. Our equation becomes

(-1)^2+2(-1)=-1=4x^3-16x-1.

4x^3-16x=0

4x(x^2-4)=4x(x-2)(x+2)=0

This equation has three solutions. They are x=-2, x=0, and x=2. We note that all of these values makes 12x^2-16 (the numerator of our expression for the derivative) finite so all of these x-values have vertical tangents. So the coordinates of the points that have vertical tangents are at (-2, -1), (0, -1), and (2, -1).

Of course, looking at an actual plot of our relation (this is NOT a function) confirms that we have the correct answer.

Original Drawing by Algebra By Hand(TM)Original Drawing by Algebra By Hand(TM)