Answer:
![f(x) = 2x+n](https://tex.z-dn.net/?f=f%28x%29%20%3D%202x%2Bn)
![f(x) = mx + 1 -2m = m(x-2) + 1](https://tex.z-dn.net/?f=f%28x%29%20%3D%20mx%20%2B%201%20-2m%20%3D%20m%28x-2%29%20%2B%201)
![\textbf{(\text{c})}](https://tex.z-dn.net/?f=%5Ctextbf%7B%28%5Ctext%7Bc%7D%29%7D)
![f(x) = 2x - 3](https://tex.z-dn.net/?f=f%28x%29%20%3D%202x%20-%203)
Step-by-step explanation:
First, we know that family of functions represents a set of functions whose equations have a similar form. In our case, a family of linear functions can be represented as
.
Now, we can take an arbitrary member of that family, a function
for some real constants
and
.
In this part of the problem, we know that
, so we consider
.
To graph several members of the family, you can plug in any real number in the equation above instead of
, since
satisfy the equation.
For
, we have ![f(x) = 2x.](https://tex.z-dn.net/?f=f%28x%29%20%3D%202x.)
For
, we have
.
For
, we have ![f(x) = 2x - 15.](https://tex.z-dn.net/?f=f%28x%29%20%3D%202x%20-%2015.)
The graphs for the values
and
are presented on the first graph below.
![\textbf{(\text{b})}](https://tex.z-dn.net/?f=%5Ctextbf%7B%28%5Ctext%7Bb%7D%29%7D)
We need to find the member of the family of linear functions such that
.
Substituting
for
in
gives
.
Now, since we have that
, we can equate
with
and express one of them in terms of the other.
![2m + n = 1 \implies n = 1 - 2m](https://tex.z-dn.net/?f=2m%20%2B%20n%20%3D%201%20%5Cimplies%20n%20%3D%201%20-%202m)
Substituting
for
in
gives the equation
![f(x) = mx + 1 -2m = m(x-2) + 1](https://tex.z-dn.net/?f=f%28x%29%20%3D%20mx%20%2B%201%20-2m%20%3D%20m%28x-2%29%20%2B%201)
which represents the wanted family. To sketch several member, we can choose any real value for
, since
satisfy the equation.
For
, we have
.
For
, we have ![f(x) = 5x - 9.](https://tex.z-dn.net/?f=f%28x%29%20%3D%205x%20-%209.)
The graph is presented below.
![\textbf{(\text{c})}](https://tex.z-dn.net/?f=%5Ctextbf%7B%28%5Ctext%7Bc%7D%29%7D)
A function belongs to both families if it satisfies both conditions; Its slope must be equal to
and
.
Let's consider a function
![f(x) = mx + n](https://tex.z-dn.net/?f=f%28x%29%20%3D%20mx%20%2B%20n)
for some real constants
and
.
The objective is to find the numeric value of the constants
and
. Since the slope must be equal to
, we obtain that
and
.
To find the numeric value of
, we use the fact that
.
Substituting
for
gives
.
On the other hand, since
, we obtain that
![4 + n = f(2) = 1 \implies 4 + n = 1 \implies n = 1 - 4 \implies n = -3](https://tex.z-dn.net/?f=4%20%2B%20n%20%3D%20f%282%29%20%3D%201%20%5Cimplies%204%20%2B%20n%20%3D%201%20%5Cimplies%20n%20%3D%201%20-%204%20%5Cimplies%20n%20%3D%20-3)
Therefore, a function that belongs to both families is
.