1. **State the problem:** Find the zeros of the function $$f(x) = x^4 - x^3 - 12x^2$$ and sketch its graph. 2. **Factor the function:** Start by factoring out the greatest common factor (GCF) from all terms. $$f(x) = x^2(x^2 - x - 12)$$ 3. **Factor the quadratic:** Factor the quadratic expression inside the parentheses. $$x^2 - x - 12 = (x - 4)(x + 3)$$ 4. **Write the fully factored form:** $$f(x) = x^2 (x - 4)(x + 3)$$ 5. **Find the zeros:** Set each factor equal to zero and solve for $$x$$. - $$x^2 = 0 \implies x = 0$$ (with multiplicity 2) - $$x - 4 = 0 \implies x = 4$$ - $$x + 3 = 0 \implies x = -3$$ 6. **Interpret multiplicity:** The zero at $$x=0$$ has multiplicity 2, meaning the graph touches the x-axis and bounces off at this point. 7. **Summary of zeros:** - $$x = 0$$ (multiplicity 2) - $$x = 4$$ - $$x = -3$$ 8. **Graph behavior:** - The graph crosses the x-axis at $$x = -3$$ and $$x = 4$$. - The graph touches and bounces off the x-axis at $$x = 0$$. - Since the leading term $$x^4$$ is positive, the ends of the graph rise to positive infinity. **Final answer:** The zeros of $$f(x)$$ are $$x = 0$$ (multiplicity 2), $$x = 4$$, and $$x = -3$$.