olynomial Question 18, 2.3.51 HW Score: $76.67 \%, 16.1$ of 21 Graphs Part 3 of 7 points Points: 0.14 of 1 For the polynomial function $f(x)=-6 x^{3}+9 x+x^{5}$, answer the parts a through e. A. The graph of $f(x)$ falls to the left and rises to the right. B. The graph of $f(x)$ rises to the left and falls to the right. C. The graph of $f(x)$ rises to the left and rises to the right. D. The graph of $f(x)$ falls to the left and falls to the right. b. Find the $x$-intercept(s). State whether the graph crosses the $x$-axis, or touches the $x$-axis and turns around each intercept. The $x$-intercept(s) is/are $0,-\sqrt{3}, \sqrt{3}$ (Type an exact answer, using radicals as needed. Use a comma to separate answers as needed. Type each only once.) At which $x$-intercept(s) does the graph cross the $x$-axis? Select the correct choice below and, if necessary, fill answer box to complete your choice. A. The $x$-intercept(s) at which the graph crosses the $x$-axis is/are $\square$. (Type an exact answer, using radicals as needed. Use a comma to separate answers as needed. Typ each answer only once.) B. There are no $x$-intercepts at which the graph crosses the $x$-axis.

Step 1 ：The leading term of the polynomial is \(x^5\), which is an odd degree with a positive coefficient. Therefore, as \(x\) approaches negative infinity, \(f(x)\) approaches negative infinity, and as \(x\) approaches positive infinity, \(f(x)\) approaches positive infinity. This means the graph of \(f(x)\) falls to the left and rises to the right.

Step 2 ：To find the \(x\)-intercepts, we set \(f(x)\) equal to zero and solve for \(x\): \(-6x^3 + 9x + x^5 = 0\).

Step 3 ：This simplifies to: \(x^5 - 6x^3 + 9x = 0\).

Step 4 ：We can factor out an \(x\): \(x(x^4 - 6x^2 + 9) = 0\).

Step 5 ：Setting each factor equal to zero gives us the solutions: \(x = 0\) and \(x^4 - 6x^2 + 9 = 0\).

Step 6 ：The second equation is a quadratic in terms of \(x^2\). Let \(y = x^2\), then we have: \(y^2 - 6y + 9 = 0\).

Step 7 ：This factors to: \((y - 3)^2 = 0\). So \(y = 3\), and substituting \(y = x^2\) back in gives \(x^2 = 3\), so \(x = \sqrt{3}, -\sqrt{3}\).

Step 8 ：Therefore, the \(x\)-intercepts are \(0, -\sqrt{3}, \sqrt{3}\).

Step 9 ：The graph crosses the \(x\)-axis at these points because the multiplicity of each root is 1, which means the graph will cross the \(x\)-axis at these points rather than touching it and turning around.

Step 10 ：\(\boxed{\text{The } x\text{-intercept(s) at which the graph crosses the } x\text{-axis is/are } 0, -\sqrt{3}, \sqrt{3}}\)