Chapter 9, Problem 24MCQ

To determine

To state: Graph the function and state domain and range.

Answer to Problem 24MCQ

Domain is $\left\{x=|x\ne 3\right\}$ and range is $\left\{f\left(x\right)|f\left(x\right)\ne 2\right\}$ .

Explanation of Solution

Given information:

The given function is $f\left(x\right)=-\frac{1}{x-3}+2$ .

Calculation:

For the function $f\left(x\right)=\frac{a}{x-h}+k$ .

The vertical asymptotes is at $x=h$ . And

The horizontal asymptotes is at $y=k$ .

The domain of the function is all real number except the value at the vertical asymptotes and the range of the function is all real number except the value at the horizontal asymptotes.

The graph of the function $f\left(x\right)=-\frac{1}{x-3}+2$ is drawn below:

  

From the above graph:

  $a=-1$ : The graph is reflected through the x axis.

  $h=3$ : The graph is translated 3 units right. There is a vertical asymptotes at $x=3$ .

  $k=2$ : The graph is translated 2 units up. There is a horizontal asymptotes at $f\left(x\right)=2$ .

Therefore,

Domain: $\left\{x=|x\ne 3\right\}$ and

Range : $\left\{f\left(x\right)|f\left(x\right)\ne 2\right\}$

Hence,

Domain is $\left\{x=|x\ne 3\right\}$ and range is $\left\{f\left(x\right)|f\left(x\right)\ne 2\right\}$ .