Cubic Function 46Ba47

1. The problem asks to find an equation for the graphed function, which is a decreasing S-shaped cubic curve. 2. Such curves are typically modeled by cubic functions of the form $$y = ax^3 + bx^2 + cx + d$$. 3. From the graph description: - The curve crosses the y-axis at about $$-1$$, so $$d = -1$$. - The curve flattens around $$(1, -1)$$, indicating a possible inflection or critical point. - The top-left portion near $$(-1, 5)$$ and bottom-right near $$(2.5, -5)$$ suggest the cubic has a negative leading coefficient (since it decreases overall). 4. Using these points, we can set up equations: - At $$x=0$$, $$y=d=-1$$. - At $$x=1$$, $$y = a(1)^3 + b(1)^2 + c(1) + d = a + b + c -1 = -1$$, so $$a + b + c = 0$$. - At $$x=-1$$, $$y = -a + b - c -1 = 5$$, so $$-a + b - c = 6$$. - At $$x=2.5$$, $$y = a(2.5)^3 + b(2.5)^2 + c(2.5) -1 = -5$$. 5. Calculate $$2.5^3 = 15.625$$ and $$2.5^2 = 6.25$$, so: $$15.625a + 6.25b + 2.5c -1 = -5$$ $$15.625a + 6.25b + 2.5c = -4$$. 6. Now solve the system: $$a + b + c = 0$$ $$-a + b - c = 6$$ $$15.625a + 6.25b + 2.5c = -4$$ 7. Add the first two equations: $$(a + b + c) + (-a + b - c) = 0 + 6$$ $$2b = 6 \\ b = 3$$ 8. Substitute $$b=3$$ into the first equation: $$a + 3 + c = 0 \\ a + c = -3$$ 9. Substitute $$b=3$$ into the third equation: $$15.625a + 6.25(3) + 2.5c = -4 \\ 15.625a + 18.75 + 2.5c = -4 \\ 15.625a + 2.5c = -22.75$$ 10. From step 8, $$c = -3 - a$$. Substitute into step 9: $$15.625a + 2.5(-3 - a) = -22.75$$ $$15.625a - 7.5 - 2.5a = -22.75$$ $$(15.625 - 2.5)a = -22.75 + 7.5$$ $$13.125a = -15.25$$ $$a = \frac{-15.25}{13.125} \approx -1.162$$ 11. Then $$c = -3 - (-1.162) = -3 + 1.162 = -1.838$$ 12. The cubic function is approximately: $$y = -1.162x^3 + 3x^2 - 1.838x - 1$$ 13. This matches the graph's shape: decreasing S-curve, crossing y-axis at -1, flattening near (1, -1), and passing near the given points. **Final answer:** $$y = -1.162x^3 + 3x^2 - 1.838x - 1$$