In the previous part, we obtained dy dx = 3t2 − 27 −2t . Next, find the points where the tangent to the curve is horizontal. (Enter your answers as a comma-separated list of ordered pairs.) x = t^3 - 3t, y = t^2 - 4

(27.55, 7.22), (-11.3, 3.21).

Step-by-step explanation:

When is the tangent to the curve horizontal?

The tangent curve is horizontal when the derivative is zero.

The derivative is:

Solving a quadratic equation:

Given a second order polynomial expressed by the following equation:

.

This polynomial has roots such that , given by the following formulas:

In this question:

So

Then

So

Enter your answers as a comma-separated list of ordered pairs.

We found values of t, now we have to replace in the equations for x and y.

t = 3.35

The first point is (27.55, 7.22)

t = -2.685

The second point is (-11.3, 3.21).

the solution is 12x-6y