1. We are asked to find the equations of 5 parabolas in the form $$f(x) = a(x-p)^2 + q$$ where $$a$$ is one of $$\frac{10}{9}, -2, -\frac{1}{2}, \frac{1}{4}$$. 2. The form $$f(x) = a(x-p)^2 + q$$ means the parabola has vertex at $$(p,q)$$ and opens upward if $$a > 0$$ or downward if $$a < 0$$. 3. For each parabola, we use the given points to find $$p$$ and $$q$$ (the vertex), then determine $$a$$ by substituting a point on the parabola. --- **Parabola A:** - Opens upward (so $$a > 0$$). - Points: $$(-6,-6), (-3,2), (-1,-4)$$. 4. Estimate vertex from points: The lowest point is $$(-1,-4)$$, so vertex $$p = -1$$, $$q = -4$$. 5. Use point $$(-3,2)$$ to find $$a$$: $$ 2 = a(-3 + 1)^2 - 4 = a(-2)^2 - 4 = 4a - 4 $$ Add 4 to both sides: $$ 2 + 4 = 4a 6 = 4a $$ Divide both sides by 4: $$ \cancel{6} = \cancel{4}a \frac{6}{4} = a $$ Simplify: $$ a = \frac{3}{2}$$ 6. Since $$a = \frac{3}{2}$$ is not in the options, check next closest option $$\frac{10}{9}$$. 7. Check if $$a = \frac{10}{9}$$ fits point $$(-3,2)$$: $$ 2 \stackrel{?}{=} \frac{10}{9}(-3 + 1)^2 - 4 = \frac{10}{9} \times 4 - 4 = \frac{40}{9} - 4 = \frac{40}{9} - \frac{36}{9} = \frac{4}{9} $$ No, so $$a = \frac{10}{9}$$ is not a good fit. 8. Since none of the options fit perfectly, choose $$a = \frac{10}{9}$$ as closest positive value. Equation parabola A: $$ f(x) = \frac{10}{9}(x + 1)^2 - 4 $$ --- **Parabola B:** - Opens downward (so $$a < 0$$). - Points: $$(-4,6), (-2,2), (0,-6)$$. 9. Vertex is highest point, likely $$(-4,6)$$, so $$p = -4$$, $$q = 6$$. 10. Use point $$(-2,2)$$ to find $$a$$: $$ 2 = a(-2 + 4)^2 + 6 = a(2)^2 + 6 = 4a + 6 $$ Subtract 6: $$ 2 - 6 = 4a -4 = 4a $$ Divide by 4: $$ \cancel{-4} = \cancel{4}a -1 = a $$ 11. $$a = -1$$ is not in options, but closest is $$-2$$ or $$-\frac{1}{2}$$. 12. Check $$a = -2$$: $$ 2 \stackrel{?}{=} -2(2)^2 + 6 = -2 \times 4 + 6 = -8 + 6 = -2 $$ No. 13. Check $$a = -\frac{1}{2}$$: $$ 2 \stackrel{?}{=} -\frac{1}{2} \times 4 + 6 = -2 + 6 = 4 $$ No. 14. So $$a = -1$$ is best fit but not in options; choose $$a = -2$$ as closest. Equation parabola B: $$ f(x) = -2(x + 4)^2 + 6 $$ --- **Parabola C:** - Opens upward (so $$a > 0$$). - Points: $$(3,2), (5,6), (1,0)$$. 15. Vertex likely at $$(3,2)$$. 16. Use point $$(5,6)$$ to find $$a$$: $$ 6 = a(5 - 3)^2 + 2 = a(2)^2 + 2 = 4a + 2 $$ Subtract 2: $$ 6 - 2 = 4a 4 = 4a $$ Divide by 4: $$ \cancel{4} = \cancel{4}a 1 = a $$ 17. $$a = 1$$ not in options, closest is $$\frac{10}{9}$$ or $$\frac{1}{4}$$. 18. Check $$a = \frac{10}{9}$$: $$ 6 \stackrel{?}{=} \frac{10}{9} \times 4 + 2 = \frac{40}{9} + 2 = \frac{40}{9} + \frac{18}{9} = \frac{58}{9} \approx 6.44 $$ Close. Equation parabola C: $$ f(x) = \frac{10}{9}(x - 3)^2 + 2 $$ --- **Parabola D:** - Opens upward. - Points: $$(1,-6), (3,2), (5,0)$$. 19. Vertex likely at $$(1,-6)$$. 20. Use point $$(3,2)$$ to find $$a$$: $$ 2 = a(3 - 1)^2 - 6 = a(2)^2 - 6 = 4a - 6 $$ Add 6: $$ 2 + 6 = 4a 8 = 4a $$ Divide by 4: $$ \cancel{8} = \cancel{4}a 2 = a $$ 21. $$a = 2$$ not in options, closest is $$\frac{10}{9}$$ or $$\frac{1}{4}$$. 22. Check $$a = \frac{10}{9}$$: $$ 2 \stackrel{?}{=} \frac{10}{9} \times 4 - 6 = \frac{40}{9} - 6 = \frac{40}{9} - \frac{54}{9} = -\frac{14}{9} \approx -1.56 $$ No. 23. Check $$a = \frac{1}{4}$$: $$ 2 \stackrel{?}{=} \frac{1}{4} \times 4 - 6 = 1 - 6 = -5 $$ No. 24. So no exact match; choose $$a = \frac{10}{9}$$ as closest. Equation parabola D: $$ f(x) = \frac{10}{9}(x - 1)^2 - 6 $$ --- **Parabola E:** - Opens downward. - Points: $$(5,0), (4,-3), (3,-4)$$. 25. Vertex likely at $$(5,0)$$. 26. Use point $$(4,-3)$$ to find $$a$$: $$ -3 = a(4 - 5)^2 + 0 = a(-1)^2 = a $$ So $$a = -3$$. 27. $$a = -3$$ not in options, closest is $$-2$$ or $$-\frac{1}{2}$$. 28. Choose $$a = -2$$ as closest. Equation parabola E: $$ f(x) = -2(x - 5)^2 + 0 = -2(x - 5)^2 $$