12. Multiple Choice 1 point Identify the graph of the following polynomial \( f(x)=-\frac{1}{4} x(x-2)(x+2)^{2} \)

We wish to identify the graph of   f(x) = –(1/4)·x·(x – 2)·(x + 2)². Step 1. Find the degree and end behavior.  • The factors are x (degree 1), (x – 2) (degree 1) and (x + 2)² (degree 2); so the overall degree is 1 + 1 + 2 = 4.  • The leading term comes from multiplying the highest‐degree term from each factor. That is, from   x · x · x² = x⁴  and multiplied by the constant –1/4, so the leading term is –(1/4)x⁴.  • Because the degree is even and the leading coefficient is negative, as x → ±∞ we have f(x) → –∞. Step 2. Locate the zeros and note their multiplicities.  • Zero at x = 0 from the factor x; multiplicity 1.  • Zero at x = 2 from (x – 2); multiplicity 1.  • Zero at x = –2 from (x + 2)²; multiplicity 2. Recall that a zero of odd multiplicity shows the graph crossing the x‐axis, while a zero of even multiplicity causes the graph to “touch” or “bounce” off the x‐axis.  • Thus the graph crosses at x = 0 and x = 2, and it touches (does not cross) at x = –2. Step 3. What does the graph look like?  • Since f is a quartic with negative leading coefficient, both ends of the graph go down.  • There are three zeros: at –2 (with a bounce), 0, and 2.  • The fact that the graph crosses at x=0 and x=2 and only touches at x=–2 (and the overall “downward‐opening” nature) helps us pick out the correct graph from a list of choices. A correct answer description is:   “The graph of f(x) is a fourth‐degree polynomial with zeros at x = –2, 0, and 2 (with the graph only touching the axis at x = –2). Since the leading coefficient is negative, the ends of the graph both fall to –∞.” Any answer that matches this description is correct.