Question

(a) Find an equation for the family of linear functions with slope 2 and sketch several members of the family.
(b) Find an equation for the family of linear functions such that (2) = 1 and sketch several members ofthe family.
(c) Which function belongs to both families?

Answer 1

Answer:

$\textbf{(\text{a})}$ $f(x) = 2x+n$

$\textbf{(\text{b})}$ $f(x) = mx + 1 -2m = m(x-2) + 1$

$\textbf{(\text{c})}$$f(x) = 2x - 3$

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

                                   $\{ ax + b | a, b \in \mathbb{R} \}$.

Now, we can take an arbitrary member of that family, a function

                                     $f(x) = mx + n$

for some real constants $m$ and $n$.

$\textbf{(\text{a})}$  

In this part of the problem, we know that $m = 2$, so we consider

                                    $f(x) = 2x +n$.

To graph several members of the family, you can plug in any real number in the equation above instead of $n$, since $\forall n \in \mathbb{R}$ satisfy the equation.

For $n = 0$, we have $f(x) = 2x.$

For $n = 15$, we have $f(x) = 2x + 15$.

For $n = -15$, we have $f(x) = 2x - 15.$

The graphs for the values $n = 0, n = 15$ and $n = -15$ are presented on the first graph below.

$\textbf{(\text{b})}$

We need to find the member of the family of linear functions such that

                                   $f(2) = 1$.

Substituting $2$ for $x$ in $f(x) = mx + n$ gives

                                 $f(2) = 2m+n$.

Now, since we have that $f(2) = 1$, we can equate $1$ with $2m +n$ and express one of them in terms of the other.  

                        $2m + n = 1 \implies n = 1 - 2m$

Substituting $1-2m$ for $n$ in $f(x) = mx + n$ gives the equation

                    $f(x) = mx + 1 -2m = m(x-2) + 1$

which represents the wanted family. To sketch several member, we can choose any real value for $m$, since $\forall m \in \mathbb{R}$ satisfy the equation.  

For $m = 0$, we have $f(x) = 1$.

For $m = 5$, we have $f(x) = 5x - 9.$

The graph is presented below.

$\textbf{(\text{c})}$

A function belongs to both families if it satisfies both conditions; Its slope must be equal to $2$ and $f(2) = 1$.

Let's consider a function

                                $f(x) = mx + n$

for some real constants $m$ and $n$.

The objective is to find the numeric value of the constants $m$ and $n$. Since the slope must be equal to $2$, we obtain that $m = 2$ and

                               $f(x) = 2x + n$.

To find the numeric value of $n$, we use the fact that $f(2) = 1$.

Substituting $2$ for $x$ gives

                              $f(2) = 2 \cdot 2 + n = 4 + n$.

On the other hand, since $f(2) = 1$, we obtain that

        $4 + n = f(2) = 1 \implies 4 + n = 1 \implies n = 1 - 4 \implies n = -3$

Therefore, a function that belongs to both families is

                                      $f(x) = 2x - 3$ .