Question

Determine the point(s), if any, at which the graph of the function has a horizontal tangent line. $$ y=x^{3}+3 x^{2} $$

Short answer

The function \(y = x^3 + 3x^2\) has horizontal tangents at the points where \(x = 0\) and \(x = -2\).

Step-by-step solution

Compute the derivative

The derivative of the function \(y = x^3 + 3x^2\) can be obtained using the power rule which says that the derivative of \(x^n\) is \(nx^{n-1}\). Applying this rule, the derivative \(y' = 3x^2 + 6x\).

Set the derivative equal to zero

Set the derivative equal to zero and solve for x to find the values of x where there is a horizontal tangent line to the function. This gives the equation \(3x^2 + 6x = 0\).

Solve for x

Factor out a 3x, leaving the equation in the form \(3x(x + 2) = 0\). Setting each factor equal to zero gives the solutions \(x = 0\) and \(x = -2\). These are the x-values where the function has a horizontal tangent.