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
