1. **Problem Statement:** We need to find a cubic function $f(x)$ that matches the described graph. 2. **Understanding the graph:** The graph is cubic, passing through the origin $(0,0)$, with two turning points: a local minimum near $x=-4$ and a local maximum near $x=4$. The function starts high at $x=-10$ and ends low at $x=10$, indicating a negative leading coefficient. 3. **General form of a cubic function:** $$f(x) = ax^3 + bx^2 + cx + d$$ Since the graph passes through the origin, $d=0$. 4. **Using turning points:** The derivative $f'(x) = 3ax^2 + 2bx + c$ has roots at the turning points $x=-4$ and $x=4$. 5. **Form derivative with roots:** $$f'(x) = k(x+4)(x-4) = k(x^2 - 16)$$ Expanding: $$f'(x) = kx^2 - 16k$$ Matching coefficients: $$3a = k, \quad 2b = 0, \quad c = -16k$$ So, $$b=0$$ $$a = \frac{k}{3}$$ $$c = -16k$$ 6. **Integrate $f'(x)$ to get $f(x)$:** $$f(x) = \int f'(x) dx = \int (kx^2 - 16k) dx = k \frac{x^3}{3} - 16k x + C$$ Since $f(0)=0$, $C=0$. 7. **Rewrite $f(x)$:** $$f(x) = k \left( \frac{x^3}{3} - 16x \right)$$ 8. **Determine $k$ using a point:** Using the point $(4,40)$ from the graph: $$40 = k \left( \frac{4^3}{3} - 16 \times 4 \right) = k \left( \frac{64}{3} - 64 \right) = k \left( \frac{64 - 192}{3} \right) = k \left( -\frac{128}{3} \right)$$ 9. **Solve for $k$:** $$k = \frac{40}{-128/3} = 40 \times \frac{3}{-128} = -\frac{120}{128} = -\frac{15}{16}$$ 10. **Final function:** $$f(x) = -\frac{15}{16} \left( \frac{x^3}{3} - 16x \right) = -\frac{15}{48} x^3 + 15 x = -\frac{5}{16} x^3 + 15 x$$ **Answer:** $$\boxed{f(x) = -\frac{5}{16} x^3 + 15 x}$$ This function matches the graph's behavior, passing through the origin, with turning points near $x=\pm4$, and the correct shape.