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

Apr 13, 2018

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

#### Explanation:

If

${y}^{2} + 2 y = 4 {x}^{3} - 16 x - 1$

Then by implicit differentiation we have

$2 y \frac{\mathrm{dy}}{\mathrm{dx}} + 2 \frac{\mathrm{dy}}{\mathrm{dx}} = 12 {x}^{2} - 16$.

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

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{12 {x}^{2} - 16}{2 y + 2}$

The tangent line will be vertical when $\frac{\mathrm{dy}}{\mathrm{dx}}$ becomes infinite. This will happen when

$2 y + 2 = 0$.

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

${\left(- 1\right)}^{2} + 2 \left(- 1\right) = - 1 = 4 {x}^{3} - 16 x - 1$.

$4 {x}^{3} - 16 x = 0$

$4 x \left({x}^{2} - 4\right) = 4 x \left(x - 2\right) \left(x + 2\right) = 0$

This equation has three solutions. They are $x = - 2$, $x = 0$, and $x = 2$. We note that all of these values makes $12 {x}^{2} - 16$ (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 $\left(- 2 , - 1\right)$, $\left(0 , - 1\right)$, and $\left(2 , - 1\right)$.

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