1. **Problem 1:** Identify the equation that matches the given graph of a downward opening parabola with vertex near $(2,4)$ and x-intercepts near $-1$ and $5.5$. - Equation options: A) $y = -x^{2} + 3$ B) $y = -x^{3} + 3$ C) $y = x^{2} - 3$ D) $y = x^{3} - 3$ - Since the graph is parabolic (quadratic) and opens downward, its equation is a quadratic with a negative leading coefficient. - Option A matches a downward parabola. - Checking vertex: vertex of $y=-x^{2}+3$ is at $(0,3)$, which is close but given graph vertex is near $(2,4)$. - The problem states it is the graph above, so likely equation A. 2. **Problem 2:** Find the parabola with vertex at $(1, 0)$. - Vertex form is $f(x) = a(x-h)^2 + k$, where vertex is $(h,k)$. - Check options: A) $x^{2} + 1$ has vertex $(0,1)$ B) $(x-1)^2$ has vertex $(1,0)$ C) $(x+1)^2$ has vertex $(-1,0)$ D) $x^{2} - 1$ has vertex $(0,-1)$ - Correct choice is B. 3. **Problem 3:** Vertex of $y = (x - 3)^2 + 6$. - Vertex form is $y=(x - h)^2 + k$, so vertex is $(3,6)$. - Correct answer B. 4. **Problem 4:** Vertex of $y = 3(x + 2)^2 + 5$. - Vertex form: $y = a(x - h)^2 + k$, vertex $(h,k)$. - Here $h = -2$, $k=5$. - Correct D. 5. **Problem 5:** Vertex of $f(x) = x^2 - 10x + 28$. - Use $h = -b/(2a) = -(-10)/2 = 5$. - Calculate $k = f(5) = 25 - 50 + 28 = 3$. - Vertex $(5,3)$. - Correct B. 6. **Problem 6:** Vertex of $f(x) = 5x^2 + 20x + 17$. - $h = -b/(2a) = -20/(2*5) = -2$. - $k = f(-2) = 5(4) + 20(-2) + 17 = 20 - 40 + 17 = -3$. - Vertex $(-2,-3)$. - Correct A. 7. **Problem 7:** Given graph of $y=f(x)$ peaks near $y=5$ at $x=2$ and crosses $y=0$ near $x=5$. - Find $f(f(2))$. - Since $f(2) ext{ (peak) } o 5$ (from graph). - Then $f(5) = 0$ (graph crosses y=0 at x=5). - So $f(f(2)) = f(5) = 0$. - Answer A. 8. **Problem 8:** Which equation matches the graph from problem 7? - Since peak at $(2,4)$ and opens downward, use vertex form: $y = a(x - 2)^2 + 4$ with $a<0$. - Option B: $y=4 - 0.25(x - 2)^2$ fits vertex at (2,4), downward opening. - Correct B. 9. **Problem 9:** Square ABCD with area 36 and $h(x) = kx^2$, where A and B on x-axis, C and D on curve. - Side length $s = ext{sqrt}(36) = 6$. - Points A and B on x-axis at distance 6, so base length $=6$. - Height (vertical) is $h = kx^2$, so height = $6$. - If C and D are at $x=3$ (midpoint), then height $= k imes 3^2 = 9k$. - Since side = 6, $9k=6 o k=6/9=2/3$. - Answer E. 10. **Problem 10:** Parabola passes $(3,4)$, vertex $(5,-2)$. - For parabola with vertex $(h,k)$, symmetric points satisfy $f(5 + d) = f(5 - d)$. - Distance from 3 to 5 is 2, so point symmetric to $(3,4)$ is $(7,4)$. - So $(7,4)$ must be on graph. - Correct D. 11. **Problem 11:** Parabola through $(0,0)$ and $(6,0)$ with turning point $T(h,4)$. - Vertex $h$ is midpoint of roots: $h = (0 + 6)/2 = 3$, so I is false (h=2 wrong), not I. - II: Parabola values symmetric around vertex, so if passes $(1,2)$ then passes $(5,2)$. - True. - III: If parabola opens downward (vertex max), then vertex is maximum. - Since $y=4$ at vertex, and roots 0,6, parabola opens downward, so III true. - Correct answer E (II and III only). 12. **Problem 12:** Quadratic $f(x) = -x^2 - 4x + 1$, axis of symmetry $x=-2$. - Vertex $(h,k)$ with $h = -b/2a = -(-4)/(2 imes -1) = 4/-2 = -2$, correct. - Compute $k = f(-2) = -4 + 8 + 1 = 5$. - Since leading coefficient negative, vertex is maximum point. - Correct A. 13. **Problem 13:** Parabola has minimum at $(1,-2)$, and $f(5) = f(c)$. - Parabola symmetric about $x=1$, so $c$ satisfies $|c-1| = |5-1|=4$. - So $c=1 - 4 = -3$ or $c=5$. - Correct B. 14. **Problem 14:** Parabola intersects x-axis at $-1$ and $6$, vertex lies midway at $x=(6-1)/2 = 2.5$. - So vertex x-coordinate is 2.5. - From options, only C has $x=2.5$. - Correct C. 15. **Problem 15:** Parabola intersects x-axis at $-2$ and $6$, $f(8) = f(p)$. - Vertex $x = (6-2)/2 = 2$. - Distance from 8 to 2 is $6$, so $p$ satisfies $|p-2|=6$. - Possible $p=2 - 6 = -4$ or $p=8$. - Given options, $-4$ is option B. --- Final answers: 1) A 2) B 3) B 4) D 5) B 6) A 7) A 8) B 9) E 10) D 11) E 12) A 13) B 14) C 15) B