1. **Problem Statement:** We need to identify which equation matches the given graph of a cubic polynomial with three x-intercepts and two turning points. 2. **Given Options:** - $f(x) = (x + 6)(x^2 - 36)$ - $f(x) = x(x^2 - 36)$ - $f(x) = x(x^2 + 36)$ - $f(x) = (x - 6)(x^2 - 36)$ 3. **Analyze each option:** - Note that $x^2 - 36 = (x - 6)(x + 6)$, so $f(x) = (x + 6)(x^2 - 36) = (x + 6)(x - 6)(x + 6) = (x + 6)^2(x - 6)$, which has roots at $x = -6$ (multiplicity 2) and $x = 6$. - $f(x) = x(x^2 - 36) = x(x - 6)(x + 6)$ has roots at $x = -6, 0, 6$ all with multiplicity 1. - $f(x) = x(x^2 + 36)$ has roots only at $x = 0$ since $x^2 + 36$ has no real roots. - $f(x) = (x - 6)(x^2 - 36) = (x - 6)(x - 6)(x + 6) = (x - 6)^2(x + 6)$ has roots at $x = 6$ (multiplicity 2) and $x = -6$. 4. **Match with graph description:** - The graph crosses the x-axis three times, indicating three distinct real roots. - The graph starts from bottom left (negative y, negative x), rises to a peak near $x = -5$, crosses x-axis between $-2$ and $0$, has a trough near $x = 2$, and rises steeply through $x = 5$. - This behavior matches a cubic with three distinct roots at $-6, 0, 6$. 5. **Conclusion:** The correct equation is $$f(x) = x(x^2 - 36) = x(x - 6)(x + 6)$$ This matches the graph's roots and turning points. **Final answer:** $f(x) = x(x^2 - 36)$