1. Problem restatement: We need to find the value of $f(f(2))$ given the graph of $y = f(x)$ that starts near 0, rises to about 3 at $x = 2.5$, and descends to 0 at $x = 5$. 2. Step 1: Find $f(2)$ from the graph. Since the curve rises to about 3 at $x=2.5$, at $x=2$ it should be slightly less than 3, roughly around 2.5. 3. Step 2: Find $f(f(2)) = f(2.5)$ since $f(2) \approx 2.5$. 4. Step 3: From the graph, $f(2.5)$ is approximately 3. 5. Conclusion: $f(f(2)) \approx 3$, so the answer is D) 3. --- 6. Problem restatement: Which equation could be $y = f(x)$ from the previous question? 7. Step 1: We know $f(2) \approx 2.5$ and $f(2.5) \approx 3$ and the graph shape is a downward opening parabola with a vertex near $x=2$ and $y=4$ (since values at 2.5 and 2 are close to 3, vertex should be a bit higher). 8. Step 2: Consider options: - A) $y = 4 + 0.25(x - 2)^2$ opens upward (coefficient positive), so vertex is minimum. - B) $y = 4 - 0.25(x - 2)^2$ opens downward, vertex max at $(2,4)$. - C) $y = 4 + 0.25(x + 2)^2$ opens upward, vertex at $(-2,4)$. - D) $y = 4 - 0.25(x + 2)^2$ opens downward, vertex at $(-2,4)$. 9. Step 3: Since the graph rises toward $x=2$ and reaches a maximum near $y=4$, the vertex near $(2,4)$ and opening downward parabola fits option B. 10. Conclusion: The correct equation is B) $y = 4 - 0.25(x - 2)^2$. --- 11. Problem restatement: Square ABCD has area 36; points A and B lie on $x$-axis, and C and D lie on graph of $h(x) = kx^2$. Find the value of $k$. 12. Step 1: Area of square $= 36$, so side length $= \sqrt{36} = 6$. 13. Step 2: Since A and B are on $x$-axis, assume coordinates: A $(0,0)$, B $(6,0)$. Then C and D must be 6 units above x-axis at $x=0$ and $x=6$ or other points to form square. 14. Step 3: Let’s define A at $(a,0)$, B at $(a+6,0)$, so side length along $x$ is 6. 15. Step 4: Then C and D have coordinates $(a, h(a))$, $(a+6, h(a+6))$ on graph $h(x) = kx^{2}$, with side length 6 vertically. 16. Step 5: Since the square is oriented normally, vertical height = 6, so $$ h(a) = h(a+6) = 6 $$ 17. Step 6: Using $h(x) = kx^{2}$, $$ k a^{2} = 6, \quad k (a+6)^{2} = 6 $$ 18. Step 7: Set equal: $$ k a^{2} = k (a+6)^{2} \Rightarrow a^{2} = (a+6)^{2} $$ which implies $$ a^{2} = a^{2} + 12a + 36 \Rightarrow 12a + 36 = 0 \Rightarrow a = -3 $$ 19. Step 8: Compute $k$ using $k a^{2} = 6$ with $a = -3$: $$ k \times (-3)^{2} = 6 \Rightarrow 9 k = 6 \Rightarrow k = \frac{6}{9} = \frac{2}{3} $$ 20. Conclusion: $k = \frac{2}{3}$, answer is E). --- 21. Problem restatement: The parabola $y = ax^{2} + bx + c$ passes through $(3,4)$ and has vertex at $(5,-2)$. Which point must also lie on the graph? 22. Step 1: Use vertex form of parabola: $$ y = a(x - 5)^{2} - 2 $$ 23. Step 2: Plug in point $(3,4)$: $$ 4 = a(3 - 5)^{2} - 2 = a( -2)^{2} - 2 = 4a - 2 $$ 24. Step 3: Solve for $a$: $$ 4a - 2 = 4 \Rightarrow 4a = 6 \Rightarrow a = \frac{3}{2} $$ 25. Step 4: Equation is now $$ y = \frac{3}{2}(x - 5)^2 - 2 $$ 26. Step 5: Check each choice for $x=7$ to see which $y$ matches: - A) $(0, 13)$ no $x=7$. - B) $(7, -8)$ $$ y = \frac{3}{2}(7 - 5)^2 - 2 = \frac{3}{2} (2)^2 - 2 = \frac{3}{2} \times 4 - 2 = 6 - 2 = 4 $$ does not equal -8. - C) $(7, -4)$ Similar computation: $y=4$, no match. - D) $(7, 4)$ matches calculation. 27. Conclusion: The point $(7, 4)$ lies on the graph; answer is D). ---