Question

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

Answer 1

A) 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.

Will continue post with next part.

Answer 2

B) 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.
\[f(x) - f(x_{1}) = m(x - x_{1}) \]
\[f(x) - 1 = m(x - 2) \]
\[f(x) = m(x-2)+1 \]

If you check, x=2, f(x) will always equal 1, so we have found the correct function family that satisfies the given information.

Answer 3

--> If you set the m equal to a few values again, you can graph the corresponding functions that you get, same as before.

C) 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.

\[f(x) = (2)(x-2) + 1 = 2x-4+1 = 2x-3\]

Even though it says 'which functions', which somewhat implies more than one, there cannot be more than one function with slope 2 that passes through the specified point. (This would imply that the lines are parallel and still share a point, which is not possible unless they share ALL points and are the same line)

Answer 4

thank you so much