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

The coordinates of the points that have vertical tangents are at

If

Then by implicit differentiation we have

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

The tangent line will be vertical when

So the tangent line will be vertical when

Now we need to calculate the

This equation has three solutions. They are

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

By signing up, you agree to our Terms of Service and Privacy Policy

To find the points where the graph of the equation y^2 + 2y = 4x^3 - 16x - 1 is tangent to the vertical line, we need to determine the values of x that make the derivative of the equation equal to infinity.

First, we differentiate the equation with respect to x:

d/dx (y^2 + 2y) = d/dx (4x^3 - 16x - 1)

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

Next, we solve for dy/dx:

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

To find the points where the graph is tangent to the vertical line, we set dy/dx equal to infinity:

(12x^2 - 16) / (2y + 2) = ∞

Since infinity is not a real number, we can conclude that there are no points where the graph is tangent to the vertical line.

By signing up, you agree to our Terms of Service and Privacy Policy

- What is the equation of the line tangent to #f(x)=(x^3 - 4) / x# at #x=2#?
- What is the equation of the line that is normal to #f(x)=-2x^2-2e^x # at # x=-2 #?
- How do you use the definition of the derivative to find f '(x) and f ''(x) for #f(x) = 4 + 9x - x^2#?
- For what value of #x# is the slope of the tangent line to #y=x^7 + (3/x)# undefined?
- How do you find the equations for the normal line to #x^2/32+y^2/8=1# through (4,2)?

- 98% accuracy study help
- Covers math, physics, chemistry, biology, and more
- Step-by-step, in-depth guides
- Readily available 24/7