Answer parts (a)-(e) for the function shown below \( f(x)=\mathrm{x}^{3}+4 \mathrm{x}^{2}-\mathrm{x}-4 \) a. Use the leading coefficient test to determine the graph's end behavior. Which statement describes the behavior at the ends of \( \mathrm{f}(\mathrm{x})=\mathrm{x}^{3}+4 \mathrm{x}^{2}-\mathrm{x}-4 \) ? A. The graph falls to the left and rises to the right B. The graph rises to the left and falls to the right. C. The graph falls to the left and to the right D. The graph rises to the left and to the right. b. Find the x -intercepts. State whether the graph possible crosses the x -axis, or touches the x -axis and turns around, at each intercept What are the x -intercepts? x \( =\square \) (Type an integer or a decimal Use a comma to separate answers as needed.

Let's solve the problem step by step. ### Part (a): Leading Coefficient Test The function given is \[ f(x) = x^3 + 4x^2 - x - 4 \] 1. **Identify the leading term**: The leading term of the polynomial is \(x^3\). 2. **Determine the degree**: The degree of the polynomial is 3, which is odd. 3. **Identify the leading coefficient**: The leading coefficient is 1, which is positive. According to the leading coefficient test: - For an odd degree polynomial with a positive leading coefficient, the graph will fall to the left and rise to the right. Thus, the correct answer for part (a) is: **A. The graph falls to the left and rises to the right.** ### Part (b): Finding the x-intercepts To find the x-intercepts, we need to solve the equation \(f(x) = 0\): \[ x^3 + 4x^2 - x - 4 = 0 \] We can use the Rational Root Theorem to test for possible rational roots. The possible rational roots are the factors of the constant term (-4) divided by the factors of the leading coefficient (1). The possible rational roots are: \[ \pm 1, \pm 2, \pm 4 \] Let's test these values to find the x-intercepts. 1. **Testing \(x = 1\)**: \[ f(1) = 1^3 + 4(1^2) - 1 - 4 = 1 + 4 - 1 - 4 = 0 \] So, \(x = 1\) is a root. 2. **Using synthetic division to factor the polynomial**: We can divide \(f(x)\) by \(x - 1\): \[ \begin{array}{r|rrrr} 1 & 1 & 4 & -1 & -4 \\ & & 1 & 5 & 4 \\ \hline & 1 & 5 & 4 & 0 \\ \end{array} \] This gives us: \[ f(x) = (x - 1)(x^2 + 5x + 4) \] 3. **Now, we need to factor \(x^2 + 5x + 4\)**: We can find the roots using the quadratic formula: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-5 \pm \sqrt{5^2 - 4 \cdot 1 \cdot 4}}{2 \cdot 1} = \frac{-5 \pm \sqrt{25 - 16}}{2} = \frac{-5 \pm 3}{2} \] This gives us: \[ x = \frac{-2}{2} = -1 \quad \text{and} \quad x = \frac{-8}{2} = -4 \] Thus, the x-intercepts are \(x = 1\), \(x = -1\), and \(x = -4\). ### Behavior at the x-intercepts - For \(x = 1\): Since the factor is \((x - 1)\), the graph crosses the x-axis. - For \(x = -1\): Since the factor is \((x + 1)\), the graph crosses the x-axis. - For \(x = -4\): Since the factor is \((x + 4)\), the graph crosses the x-axis. ### Final Answers for Part (b) The x-intercepts are: \[ x = 1, -1, -4 \] The graph crosses the x-axis at each intercept. So, the final answer for part (b) is: \[ x = 1, -1, -4 \]