The graph of \( f(x)=x^{3}+4 x^{2} \) is concave downward on which interval(s)? \( \begin{array}{llll}\text { A. }\left(-\frac{3}{4}, \infty\right) & \text { B. }(0, \infty) & \text { C. }\left(-\infty,-\frac{4}{3}\right) & \text { D. }\left(-\frac{4}{3}, \infty\right)\end{array} \) E. \( \left(-\infty,-\frac{3}{4}\right) \)

To determine the concavity of the function \( f(x) = x^{3} + 4x^{2} \), we need to find the second derivative of the function. First, we compute the first derivative: \[ f'(x) = 3x^2 + 8x. \] Next, we find the second derivative: \[ f''(x) = 6x + 8. \] To find intervals of concavity, we set the second derivative equal to zero: \[ 6x + 8 = 0 \Rightarrow x = -\frac{4}{3}. \] Now, we analyze the sign of \( f''(x) \) around this point: - For \( x < -\frac{4}{3} \), choose \( x = -2 \): \[ f''(-2) = 6(-2) + 8 = -12 + 8 = -4 \quad (\text{concave down}) \] - For \( x > -\frac{4}{3} \), choose \( x = 0 \): \[ f''(0) = 6(0) + 8 = 8 \quad (\text{concave up}) \] Thus, the function is concave downward on the interval \( \left(-\infty, -\frac{4}{3}\right) \). The correct answer is: E. \( \left(-\infty,-\frac{3}{4}\right) \).