1. The problem asks for the turning point of the quadratic function $f(x) = x^2 - 3x + 4$ after three transformations: reflection in the x-axis, horizontal shift 2 units right, and vertical shift 3 units up. 2. The turning point (vertex) of a quadratic $ax^2 + bx + c$ is given by the formula $x = -\frac{b}{2a}$. 3. First, find the vertex of the original function: $$a = 1, b = -3$$ $$x = -\frac{-3}{2 \times 1} = \frac{3}{2} = 1.5$$ 4. Find $f(1.5)$: $$f(1.5) = (1.5)^2 - 3(1.5) + 4 = 2.25 - 4.5 + 4 = 1.75$$ So the original turning point is at $(1.5, 1.75)$. 5. Apply the transformations step-by-step to the function: - Reflection in the x-axis: $g(x) = -f(x) = -(x^2 - 3x + 4) = -x^2 + 3x - 4$ - Horizontal shift 2 units right: $h(x) = g(x - 2) = - (x - 2)^2 + 3(x - 2) - 4$ - Vertical shift 3 units up: $k(x) = h(x) + 3 = - (x - 2)^2 + 3(x - 2) - 4 + 3$ 6. Simplify $k(x)$: $$k(x) = - (x^2 - 4x + 4) + 3x - 6 - 4 + 3 = -x^2 + 4x - 4 + 3x - 6 - 4 + 3$$ $$k(x) = -x^2 + 7x - 11$$ 7. Find the vertex of $k(x)$: $$a = -1, b = 7$$ $$x = -\frac{7}{2 \times (-1)} = -\frac{7}{-2} = 3.5$$ 8. Find $k(3.5)$: $$k(3.5) = - (3.5)^2 + 7(3.5) - 11 = -12.25 + 24.5 - 11 = 1.25$$ 9. Therefore, the turning point of the transformed function is at $(3.5, 1.25)$. 10. The correct answer choice is 4) $(3.5, 1.25)$.