F(x) = 3x+2, x≤2 f(x) = -0,5x-3, -2

D(f) - domain of the function: D(f) = {x | x ≤ 2} U {-2 < x < 0} U {x ≥ 0}

E(f) - range (or image) of the function: E(f) = {f(x) | x ≤ 2} U {f(x) | -2 < x < 0} U {f(x) | x ≥ 0}

The zeros of the function (where f(x) = 0): For the first part of the function: 3x + 2 = 0 3x = -2 x = -2/3

For the second part of the function: -0.5x - 3 = 0 -0.5x = 3 x = -6

The point of intersection with the y-axis (where x = 0): For the third part of the function: f(0) = -5

To determine whether the function is increasing or decreasing, we need to examine the slopes of the segments: The slope of the first part of the function, 3x + 2, is positive, so it is increasing. The slope of the second part of the function, -0.5x - 3, is negative, so it is decreasing. The slope of the third part of the function, -5, is constant, so it neither increases nor decreases.

Therefore, the function is increasing for x ≤ -2/3, then decreasing for -2/3 < x < 0, and finally remains constant for x ≥ 0.