For the polynimial function \( f(x)=x^{4}+4 x^{3}+4 x^{2} \), answer the parts a through e a. Use the Leading Coefficient Test to determine the graph's end behavior. A. The graph of \( f(x) \) fals to the left and falls to the right. B. The graph of \( f(x) \) rises to the lefl and falls to the right C. The graph of \( 1(x) \) rises to the left and rises to the right. D. The graph of \( 1(x) \) falls to the left and rises to the right. b. Find the \( x \)-intercept(s). Stalo whither the graph crosses the \( x \)-axis, or touches the \( x \)-axis and turns around, at each intercept The \( x \)-inlercept(s) is/are \( \square \) (Type an integer ut a decinal. Use a comma to separate answers as needed Type each answer only once.)

**a. Leading Coefficient Test** The highest degree term in \[ f(x)=x^4+4x^3+4x^2 \] is \[ x^4. \] Since the degree is even (4) and the leading coefficient is positive, the graph rises on both ends. In other words, as \( x \to -\infty \), \( f(x) \to +\infty \) and as \( x \to +\infty \), \( f(x) \to +\infty \). 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** First, factor the polynomial: \[ f(x)=x^4+4x^3+4x^2 = x^2(x^2+4x+4). \] Notice that the quadratic factors further: \[ x^2+4x+4=(x+2)^2. \] Thus, the factorization becomes: \[ f(x)=x^2(x+2)^2. \] The \( x \)-intercepts occur when \( f(x)=0 \): \[ x^2(x+2)^2=0. \] Set each factor equal to zero. 1. For \( x^2=0 \): \[ x=0. \] 2. For \( (x+2)^2=0 \): \[ x+2=0 \quad \Rightarrow \quad x=-2. \] Both factors are squared, meaning each \( x \)-intercept has even multiplicity. An intercept with even multiplicity indicates that the graph touches the \( x \)-axis and turns around at that point. Thus, the \( x \)-intercepts are: \[ 0,\ -2. \]