Answer:

a) To graph f(x) = mx + (3-2m), we need to substitute different values of m and plot the resulting graphs.

For m = 0, f(x) = 0x + (3-2 * 0) = 3

For m = 1, f(x) = 1x + (3-2 * 1) = 1x + 1

For m = 2, f(x) = 2x + (3-2 * 2) = 2x - 1

For m = -3, f(x) = -3x + (3-2 * -3) = -3x + 9

For m = -5, f(x) = -5x + (3-2 * -5) = -5x + 23

b) All the graphs in part a) have the common characteristic of having the same slope, but different y-intercepts.

c) Analytically, this can be proven by noting that the slope of all the graphs is m, and the y-intercept of each graph is (3-2m), which is different for each value of m. The slope and y-intercept of a linear function completely determine its graph.

d) The equation of the family of functions whose graphs pass by the same point (-2,-4) can be found by solving for m. Let's say f(x) = mx + (b-2m). We know that when x = -2, the function must equal -4, so we can write an equation using this information:

-4 = m * (-2) + (b - 2 * m)

Expanding and solving for m and b, we get:

-4 = -2m + b

-2m = -4 - b

-2m = -4 - b

m = 2 + b/2

Substituting m back into f(x) = mx + (b-2m), we get:

f(x) = (2 + b/2) x + (b - 2 * (2 + b/2))

f(x) = 2x + b/2 + b - 2 * 2 - 2 * b/2

f(x) = 2x - 4