1. **State the problem:** We have a cubic function $f(x) = ax^3 + bx^2 + cx - 8$ with zeros at $x = -1$ and $x = 2$, and it passes through the point $(1, -2)$. We want to find the remaining zero and understand the domain and range. 2. **Use the factor theorem:** Since $x = -1$ and $x = 2$ are zeros, $(x + 1)$ and $(x - 2)$ are factors of $f(x)$. So, $f(x) = a(x + 1)(x - 2)(x - r)$ where $r$ is the remaining zero. 3. **Use the point $(1, -2)$ to find $a$ and $r$:** Substitute $x=1$ and $f(1) = -2$: $$-2 = a(1 + 1)(1 - 2)(1 - r) = a(2)(-1)(1 - r) = -2a(1 - r)$$ 4. **Simplify:** $$-2 = -2a(1 - r) \implies 1 = a(1 - r)$$ 5. **Express $a$ in terms of $r$:** $$a = \frac{1}{1 - r}$$ 6. **Use the fact that the constant term is $-8$:** The constant term of $f(x)$ is $a \times (1) \times (-2) \times (-r) = -8$. Expand the factors at $x=0$: $$f(0) = a(0 + 1)(0 - 2)(0 - r) = a(1)(-2)(-r) = 2ar = -8$$ 7. **Substitute $a$ from step 5:** $$2 \times \frac{1}{1 - r} \times r = -8 \implies \frac{2r}{1 - r} = -8$$ 8. **Solve for $r$:** $$2r = -8(1 - r) = -8 + 8r$$ $$2r - 8r = -8$$ $$-6r = -8$$ $$r = \frac{8}{6} = \frac{4}{3}$$ 9. **Find $a$:** $$a = \frac{1}{1 - \frac{4}{3}} = \frac{1}{\frac{-1}{3}} = -3$$ 10. **Final function:** $$f(x) = -3(x + 1)(x - 2)\left(x - \frac{4}{3}\right)$$ 11. **Domain:** The domain of any cubic polynomial is all real numbers, so: $$\text{Domain} = (-\infty, \infty)$$ 12. **Range:** Cubic functions have range all real numbers, so: $$\text{Range} = (-\infty, \infty)$$ 13. **Points for graph:** - Zeros: $x = -1, 2, \frac{4}{3}$ - Y-intercept: $f(0) = -8$ - Point given: $(1, -2)$ 14. **Additional points for smooth graph:** Calculate $f(-2)$: $$f(-2) = -3(-2 + 1)(-2 - 2)\left(-2 - \frac{4}{3}\right) = -3(-1)(-4)(-\frac{10}{3}) = -3 \times -1 \times -4 \times -\frac{10}{3} = -40$$ Calculate $f(3)$: $$f(3) = -3(3 + 1)(3 - 2)\left(3 - \frac{4}{3}\right) = -3(4)(1)(\frac{5}{3}) = -20$$ Calculate $f(-0.5)$: $$f(-0.5) = -3(-0.5 + 1)(-0.5 - 2)\left(-0.5 - \frac{4}{3}\right) = -3(0.5)(-2.5)(-1.8333) \approx -6.875$$