For the polynomial function \( 1(x)=x^{4}+4 x^{3}+4 x^{2} \), enswer the parts a through e. a. Use the Leading Coofficient Test to delermine 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 lelt and falls to the right C. The graph of \( f(x) \) rises to the lelt and rises to the right. D. The graph of \( f(x) \) falls to the lefl and rises to the right. b. Find the \( x \)-intercapt(5). Stato whether the graph crosses tho \( x \)-axis, or touches the \( x \)-axis and turns around, at each inter The \( x \)-intercept(s) is/are \( \square \) ]. (Type an integer or a decimal. Uso a comma lo senarale answers as needed. Type each answer only once.)

**a. Leading Coefficient Test** The polynomial is \[ f(x)=x^4+4 x^3+4 x^2. \] First, note that this polynomial is degree 4 (an even degree) and its leading coefficient is 1 (positive). The Leading Coefficient Test tells us that when the degree of a polynomial is even and the leading coefficient is positive, the graph rises on both ends. Thus, the correct answer is: **C. The graph of \( f(x) \) rises to the left and rises to the right.** --- **b. Finding the \( x \)-intercepts** To find the \( x \)-intercepts of \( f(x) \), set \( f(x)=0 \): \[ x^4+4 x^3+4 x^2=0. \] Factor out the common term \( x^2 \): \[ x^2 (x^2+4 x+4)=0. \] Recognize that the quadratic is a perfect square: \[ x^2+4x+4=(x+2)^2. \] Thus, we can write \[ x^2 (x+2)^2=0. \] Setting each factor equal to zero gives: - \( x^2=0 \) which implies \( x=0 \), - \( (x+2)^2=0 \) which implies \( x+2=0 \) or \( x=-2 \). Since both factors are squared (even multiplicity), at each \( x \)-intercept the graph touches the \( x \)-axis and turns around. Thus, the \( x \)-intercepts are: \[ x = -2 \quad \text{and} \quad x = 0. \]