1. **State the problem:** Find the quadratic equation in factored form given two x-intercepts and a point on the graph. 2. **Recall the formula:** A quadratic with roots $r_1$ and $r_2$ can be written as $$y = a(x - r_1)(x - r_2)$$ where $a$ is a constant to be determined using the given point. 3. **Use the first set:** x-intercepts $(7,0)$ and $(-8,0)$, point $(8,-16)$. 4. Write the equation with unknown $a$: $$y = a(x - 7)(x + 8)$$ 5. Substitute point $(8,-16)$: $$-16 = a(8 - 7)(8 + 8) = a(1)(16) = 16a$$ 6. Solve for $a$: $$a = \frac{-16}{16} = -1$$ 7. Final equation: $$y = -1(x - 7)(x + 8) = -(x - 7)(x + 8)$$ --- 8. **Use the second set:** x-intercepts $(4,0)$ and $(4,0)$ (a repeated root), point $(-2,12)$. 9. Equation form: $$y = a(x - 4)^2$$ 10. Substitute point $(-2,12)$: $$12 = a(-2 - 4)^2 = a(-6)^2 = 36a$$ 11. Solve for $a$: $$a = \frac{12}{36} = \frac{1}{3}$$ 12. Final equation: $$y = \frac{1}{3}(x - 4)^2$$ --- 13. **Use the third set:** x-intercepts $(\frac{2}{3},0)$ and $(-3,0)$, point $(3,14)$. 14. Equation form: $$y = a\left(x - \frac{2}{3}\right)(x + 3)$$ 15. Substitute point $(3,14)$: $$14 = a\left(3 - \frac{2}{3}\right)(3 + 3) = a\left(\frac{7}{3}\right)(6) = a \times 14$$ 16. Solve for $a$: $$a = \frac{14}{14} = 1$$ 17. Final equation: $$y = \left(x - \frac{2}{3}\right)(x + 3)$$