For the polynomial function \( f(x)=-4 x^{4}+8 x^{3} \), answer the parts a through e. a. Use the Leading Coefficient Test to determine the graph's end behavior. A. The graph of \( f(x) \) falls to the left and falls to the right. B. The graph of \( f(x) \) rises to the left and rises to the right. C. The graph of \( f(x) \) rises to the left and falls to the right. D. The graph of \( f(x) \) falls to the left and rises to the right. The the x-intercepts. State whether the graph crosses the \( x \)-axis, or touches the \( x \)-axis and turns around, at each intere \( \square \). (Type an integer or a decimal. Use a comma to separate answers as needed. Type each answer only once.) .

To determine the end behavior of the polynomial function \( f(x) = -4x^4 + 8x^3 \), we can look at the leading term, which is \( -4x^4 \). Since the leading coefficient is negative and the degree of the polynomial is even, we can conclude that the graph falls to the left and falls to the right. So the correct answer is: **A. The graph of \( f(x) \) falls to the left and falls to the right.** Next, to find the x-intercepts, we set the function equal to zero: \[ -4x^4 + 8x^3 = 0 \implies -4x^3(x - 2) = 0 \] This gives us the solutions \( x = 0 \) (with a multiplicity of 3) and \( x = 2 \) (with a multiplicity of 1). Since \( x = 0 \) has an odd multiplicity, the graph crosses the x-axis at this intercept. At \( x = 2 \), with an odd multiplicity, the graph also crosses the x-axis. So, the x-intercepts are \( 0, 2 \) and the behavior is: **The graph crosses the x-axis at \( 0 \), and crosses the x-axis at \( 2 \).**