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

1 Answer
Mr. Mike
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)

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