Question

1. state the transformations, domain/range, vertical and horizontal asymptotes. sketch rational \\( f(x) = \frac{1}{x - 5} + 2 \\)\

\\( \quad \\)a. state the transformations\
\\( \quad \\)b. domain:\
\\( \quad \\)c. range:\
\\( \quad \\)d. vertical asymptote:\
\\( \quad \\)e. horizontal asymptote:\

1. state the transformations, domain/range, vertical and horizontal asymptotes. sketch rational \\( f(x) = -\frac{1}{x + 1} - 5 \\)\

\\( \quad \\)a. state the transformations\
\\( \quad \\)b. domain:\
\\( \quad \\)c. range:\
\\( \quad \\)d. vertical asymptote:\
\\( \quad \\)e. horizontal asymptote:

Explanation:

Step1: Analyze the parent function

The parent function for rational functions like \( f(x)=\frac{1}{x} \) has a vertical asymptote at \( x = 0 \), horizontal asymptote at \( y=0 \), domain \( (-\infty,0)\cup(0,\infty) \), and range \( (-\infty,0)\cup(0,\infty) \).

Step2: Analyze transformations for \( f(x)=-\frac{1}{x + 1}-5 \)

• Horizontal shift: The \( x+1 \) in the denominator means a shift of the parent function \( \frac{1}{x} \) left by 1 unit (since for \( y=\frac{1}{x - h} \), \( h=- 1 \) here, so shift left 1).
• Reflection: The negative sign in front of \( \frac{1}{x + 1} \) reflects the graph over the \( x \)-axis.
• Vertical shift: The \( - 5 \) at the end shifts the graph down by 5 units.

Step3: Find the domain

The function is undefined when the denominator \( x + 1=0\), i.e., \( x=-1 \). So the domain is all real numbers except \( x = - 1 \), which is \( (-\infty,-1)\cup(-1,\infty) \).

Step4: Find the range

Since the horizontal asymptote is \( y=-5 \) (from the vertical shift and the nature of the rational function transformation), the range is all real numbers except \( y=-5 \), so \( (-\infty,-5)\cup(-5,\infty) \).

Step5: Find vertical asymptote

Vertical asymptote occurs where the denominator is zero, so \( x=-1 \).

Step6: Find horizontal asymptote

For rational functions of the form \( \frac{a}{x - h}+k \), the horizontal asymptote is \( y = k \). Here \( k=-5 \), so horizontal asymptote is \( y=-5 \).

Answer:

Part a (Transformations)

• Horizontal shift: Left by 1 unit.
• Reflection: Over the \( x \)-axis.
• Vertical shift: Down by 5 units.

Part b (Domain)

\( (-\infty,-1)\cup(-1,\infty) \)

Part c (Range)

\( (-\infty,-5)\cup(-5,\infty) \)

Part d (Vertical Asymptote)

\( x=-1 \)

Part e (Horizontal Asymptote)

\( y = - 5 \)