Question

let ( f(x) = \frac{1}{x + 3} + 2 ).
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 function \( f(x)=\frac{1}{x + 3}+2 \), the vertical asymptote occurs where the denominator is zero (since division by zero is undefined).
Set \( x + 3=0 \).

Step2: Solve for x

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

Step1: Recall horizontal asymptote rule for rational functions

For a function of the form \( y=\frac{a}{x - h}+k \), the horizontal asymptote is \( y = k \). Here, \( f(x)=\frac{1}{x + 3}+2 \), so \( k = 2 \).

Step1: Determine domain restrictions

The function \( f(x)=\frac{1}{x + 3}+2 \) is undefined when \( x+3 = 0\) (i.e., \( x=-3 \)). For all other real numbers, the function is defined.

Step2: Write domain in interval notation

The domain is all real numbers except \( x=-3 \), so in interval notation, it is \( (-\infty,-3)\cup(-3,\infty) \).

Answer:

\( x = - 3 \)

Part (b)