Question

let $f(x) = \frac{1}{x + 3} + 5$.
a) enter the equation of the vertical asymptote of $f(x)$.
b) enter the equation of the horizontal asymptote of $f(x)$.
c) enter the domain of $f(x)$ in interval notation.
d) enter the range of $f(x)$ in interval notation.

Explanation:

Part (a)

Step1: Recall vertical asymptote rule

For a rational function \( y = \frac{1}{x - a}+b \), vertical asymptote is at \( x = a \) (where denominator is zero).
Given \( f(x)=\frac{1}{x + 3}+5=\frac{1}{x-(-3)}+5 \), set denominator \( x + 3=0 \).

Step2: Solve for x

\( x+3 = 0\implies x=-3 \).

Step1: Recall horizontal asymptote rule

For \( y=\frac{1}{x - a}+b \), horizontal asymptote is \( y = b \) (as \( x\to\pm\infty \), \( \frac{1}{x - a}\to0 \)).
Given \( f(x)=\frac{1}{x + 3}+5 \), as \( x\to\pm\infty \), \( \frac{1}{x + 3}\to0 \), so \( f(x)\to5 \).

Step1: Identify undefined point

The function \( f(x)=\frac{1}{x + 3}+5 \) is undefined when denominator \( x + 3 = 0\implies x=-3 \).

Step2: Determine domain intervals

For all real numbers except \( x=-3 \), the function is defined. So domain is \( (-\infty,-3)\cup(-3,\infty) \).

Answer:

\( x = -3 \)

Part (b)