Final answer:
The function f(x)=2x²-4x+1 is decreasing for x < 1 and increasing for x > 1. The graph of f(x) is an upward opening parabola with a vertex at (1, f(1)) and a y-intercept at (0, 1).
Step-by-step explanation:
To determine where the function f(x)=2x²-4x+1 is increasing or decreasing, we need to find its derivative to identify the critical points. The derivative of f(x) is f'(x) = 4x - 4. Setting this equal to zero gives us the critical point at x = 1. To analyze the behavior of f(x) around this point, we use a sign chart: for x < 1, f'(x) is negative, and for x > 1, f'(x) is positive. Therefore, f(x) is decreasing on the interval (-∞, 1) and increasing on the interval (1, ∞). For part (b), to sketch a graph of y=f(x), we follow the behavior identified: the graph is a parabola opening upwards (since the coefficient of x² is positive), with a vertex at (1,f(1)). The y-intercept is f(0) = 1, and the function decreases to its minimum at x = 1 before increasing afterward.