Chapter 4.6, Problem 84E

To determine

To Find: To find the domain, intercepts, and asymptotes of the given function.

Answer to Problem 84E

Domain is $R-\{-2\}$

Intercepts of x and y are $7\text{and}\frac{-7}{4}$ and

Asymptotes of vertical and horizontal are $x=-2\text{and}y=0.$

Explanation of Solution

Given information: Given $f(x)=\frac{x-7}{x^2+4x+4}$

Calculation:

The given function is $f(x)=\frac{x-7}{x^2+4x+4}$

For calculating domain, we have

  $\begin{array}{l}{x}^2+4x+4\ne 0\\ {(}^x\ne 0\\ (x+2)\ne 0\\ x\ne -2\end{array}$

Thus, the domain of the given function is $R-\{-2\}$ .

For calculating intercepts, we have

  $\begin{array}{l}\text{For}x\text{-intercept}\text{we}\text{put}f(x)=0,\text{then}\\ \frac{x-7}{x^2+4x+4}=0\\ x-7=0\\ x=7\\ \text{and}\text{for}y\text{-intercept}\text{we}\text{put}x=0,\text{then}\\ y=\frac{0-7}{0^2+4\times 0+4}\\ y=\frac{-7}{4}\end{array}$

Thus, the required intercept for $x\text{and}y\text{is}7\text{and}\frac{-7}{4}.$

For calculating asymptotes, we have

For vertical asymptotes, we put

  $\begin{array}{l}{x}^2+4x+4=0\\ $ {(x+2)}^2=0$x+2=0\\ \therefore x=-2\end{array}$

And for horizontal asymptotes, we have

  $\begin{array}{l}\because \text{degree}\text{of}\text{numerator}<\text{degree}\text{of}\text{denominator}\\ \therefore y=0.\end{array}$

Thus, the vertical and horizontal asymptotes are $x=-2\text{and}y=0.$