Use a graphing utility to locate all horizontal tangent lines to the curve y = x³ + 3x² + 12. A) x = 0 B) x = -3 and x = 0 C) x = -2 and x = 0 D) x = -2 E) x = -3

Question

Quizplus · Accepted Answer

First, take the derivative of the curve: y' = 13x² + 6x¹ Set y' equal to zero, since we are looking for horizontal tangent lines: 13x² + 6x¹ = 0 x(13x¹ + 6) = 0 Solve for x: x = 0 or x = -6/13 Next, plot the curve and use the trace function to find the y-value of the curve at each value of x. At x = 0, the y-value is 12. This confirms that there is a horizontal tangent line at x = 0. At x = -6/13, the y-value is approximately 1.62. There is no horizontal tangent line at this x-value. Therefore, the best choice is C, as it includes the x-value of 0, which we have already confirmed has a horizontal tangent line.

Use a Graphing Utility to Locate All Horizontal Tangent Lines