Originally Posted by

**MarkFL2** As mentioned, the slope of the line y = -3 is zero, thus the slope of the function described by the differential equation will be zero when y = -3. So set your derivative equal to zero and let y = -3 and solve for x. (As you will see, you really don't need to know what y is...)

Then, to determine the nature of the function's behavior at that point, use the second derivative test, using the quotient and chain rules, and substitute into your formula for y'' the values for x, y, and y' to find the sign of y'' at the tangent point.