1. The problem is to find the zeros of the function $$f(x) = x^4 - x^3 - 12x^2$$ and identify the points where the graph crosses the x-axis. 2. The function can be factored as $$f(x) = x^2(x^2 - x - 12)$$. 3. Further factor the quadratic inside the parentheses: $$x^2 - x - 12 = (x - 4)(x + 3)$$ 4. So the full factorization is: $$f(x) = x^2 (x - 4)(x + 3)$$ 5. The zeros of the function are the values of $x$ that make each factor zero: - From $x^2 = 0$, we get $x = 0$ with multiplicity 2. - From $x - 4 = 0$, we get $x = 4$. - From $x + 3 = 0$, we get $x = -3$. 6. The points where the graph crosses the x-axis are therefore: $$ (0, 0), (4, 0), (-3, 0) $$ 7. Since $x=0$ is a root of multiplicity 2, the graph touches the x-axis at $(0,0)$ and turns around there. 8. To find the y-values for other points or to sketch the graph, note the behavior: - The graph dips deeply below the x-axis near $x = -3$. - The ends of the graph go to positive infinity as $x \to \pm \infty$ because the leading term $x^4$ dominates. Final answer: The zeros and points on the graph where it crosses the x-axis are $$(-3, 0), (0, 0), (4, 0)$$.