For Part (a), we are given the function f(x) = 2x² – 5x + 3 and asked to find the point where the tangent line to this function is horizontal. A horizontal tangent line occurs where the slope of the function is zero. Since the derivative of a function gives the slope at any point, we need to find where f'(x) = 0. First, we compute the derivative: f'(x) = 4x – 5. Setting this equal to zero gives 4x – 5 = 0, which solves to x = 5/4. This is the x-coordinate of the point where the tangent is horizontal. To find the corresponding y-coordinate, we substitute x = 5/4 back into the original function: f(5/4) = 2(5/4)² – 5(5/4) + 3. Simplifying step by step: (5/4)² = 25/16, so 2 × 25/16 = 50/16 = 25/8. Next, -5 × (5/4) = -25/4. To combine these terms, we express everything with a common denominator of 8: 25/8 – 50/8 + 24/8 = (25 – 50 + 24)/8 = (-1)/8. Therefore, f(5/4) = -1/8. Hence, the point where the tangent line is horizontal is (5/4, -1/8).