Chapter 2.6, Problem 34E

Solution

The match of the functions, $f\left(x\right)=1+\frac{1}{x+3}$ with given graphs.

The graph of $f\left(x\right)=1+\frac{1}{x+3}$ is (f).

Given:

The graphs are given as follows,

  

  

Concept Used:

The graph of a function of type $f\left(x\right)=\frac{ax+b}{cx+d}$ where $c\ne 0$ can be obtained by using the transformation of basic reciprocal function $\frac{1}{x}$ .

The graph of basic reciprocal function $f\left(x\right)=\frac{1}{x}$ is given by,

  

Here, horizontal asymptotes is $x=0$ , vertical asymptotes is, $y=0$ and $\underset{x\to {\infty }^{-}}{\lim }f\left(x\right)=\underset{x\to {\infty }^{+}}{\lim }f\left(x\right)=0$ .

And, by translation property, the of graph $a+f\left(x\right)$ is the vertical translation of $f\left(x\right)$ by a units up or down according a is positive or negative respectively.

And, the graph of $f\left(x+a\right)$ is the horizontal translation of $f\left(x\right)$ by a units left.

Calculation:

Consider the function, $f\left(x\right)=1+\frac{1}{x+3}$ .

If the basic reciprocal function is $g\left(x\right)=\frac{1}{x}$ , then $f\left(x\right)=1+g\left(x+3\right)$ is the vertical translation by one unit up and horizontal translation by three units left.

The graph of $f\left(x\right)=1+\frac{1}{x+3}$ is as follows,

  

Hence, from the given graphs, the graph of $f\left(x\right)=1+\frac{1}{x+3}$ is (f).

Conclusion:

The graph of $f\left(x\right)=1+\frac{1}{x+3}$ is (f).