A) find an equation for the family of linear functions with slope 2 and sketch several members of the family
b)a) find an equation for the family of linear functions such that f(2) =1 and sketch several members of the family
c)which functions belongs to both families

They are asking us to find another 'family' of linear functions. This time, they are asking that 'f(2) = 1' be true for any f(x) in the family. Because we are now provided with a point and no slope, it would be best to use a point-slope form definition. (If we use slope-intercept, we would end up with an unknown intercept and nothing to do about it!) So, our point would be (2,1). The slope will remain as a variable. f(x)−f(x1)=m(x−x1) f(x)−1=m(x−2) f(x)=m(x−2)+1 If you check, x=2, f(x) will always equal 1, so we have found the correct function family that satisfies the given information.They want you to define a 'family' of linear functions and they give you only a slope, so we are going to have a set of functions in the form f(x)=mx+b;m=2 f(x)=2x+b This function, "f(x) = 2x + b," can represent any linear function given some y-intercept 'b', so this will be sufficient for first part of our answer. Then, we are asked to graph a few functions of the family. So, we can choose any values for 'b' that we want and graph these on the xy-plane. I would recommend choosing b values close to 0 for ease of graphing. So, b=1, b=0, and b=-1 should be sufficient (they really are not clear about 'how many'). I assume you know how to graph them, but I would be glad to show you how if you are not certain.

We know that to meet the requirements of BOTH parts, it has to be a function of slope 2 and value of f(2) = 1. Well, we already have a function that defines the f(2)=1 is true part, and if we substitute m=2 into that family, then we will have a function of slope 2. f(x)=(2)(x−2)+1=2x−4+1=2x−3