Question

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

1. state the transformations, domain/range, vertical and horizontal asymptotes. sketch rational

f(x) = -\frac{1}{x + 1} - 5
a. state the transformations
b. domain
c. range
d. vertical asymptote:
e. horizontal asymptote:

Explanation:

For \( f(x)=\frac{1}{x - 5}+2 \)

Step 1: Transformations

The parent function is \( y=\frac{1}{x} \). For the function \( y=\frac{1}{x - h}+k \), when \( h = 5 \) and \( k=2 \), the graph of \( y = \frac{1}{x} \) is shifted horizontally by \( h \) units (right if \( h>0 \)) and vertically by \( k \) units (up if \( k > 0 \)). So, it is a horizontal shift right by 5 units and a vertical shift up by 2 units.

Step 2: Domain

The function is undefined when the denominator \( x - 5=0\), i.e., \( x = 5 \). So the domain is all real numbers except \( x = 5 \), which is \( (-\infty,5)\cup(5,\infty) \).

Step 3: Range

As \( x\to\pm\infty \), \( \frac{1}{x - 5}\to0 \), so \( f(x)\to2 \). But \( f(x) \) can never be equal to 2. Also, as \( x\to5^{+} \) or \( x\to5^{-} \), \( \frac{1}{x - 5}\to\pm\infty \), so \( f(x)\to\pm\infty \). Thus, the range is \( (-\infty,2)\cup(2,\infty) \).

Step 4: Vertical Asymptote

The vertical asymptote occurs where the denominator is zero, so \( x-5 = 0\Rightarrow x = 5 \).

Step 5: Horizontal Asymptote

For rational functions of the form \( y=\frac{a}{x - h}+k \) (where \( a
eq0 \)), the horizontal asymptote is \( y = k \). Here \( k = 2 \), so the horizontal asymptote is \( y=2 \).

For \( f(x)=-\frac{1}{x + 1}-5 \)

Step 1: Transformations

The parent function is \( y=\frac{1}{x} \). The function \( y=-\frac{1}{x+1}-5 \) can be seen as a transformation of \( y = \frac{1}{x} \). The \( x+1 \) inside the denominator means a horizontal shift left by 1 unit (since \( y=\frac{1}{x - (-1)} \)). The negative sign in front of \( \frac{1}{x + 1} \) is a reflection over the x - axis. The \( - 5 \) at the end is a vertical shift down by 5 units.

Step 2: Domain

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

Step 3: Range

As \( x\to\pm\infty \), \( -\frac{1}{x + 1}\to0 \), so \( f(x)\to - 5 \). But \( f(x) \) can never be equal to - 5. Also, as \( x\to - 1^{+} \) or \( x\to - 1^{-} \), \( -\frac{1}{x + 1}\to\pm\infty \), so \( f(x)\to\pm\infty \). Thus, the range is \( (-\infty,-5)\cup(-5,\infty) \).

Step 4: Vertical Asymptote

The vertical asymptote occurs where the denominator is zero, so \( x + 1=0\Rightarrow x=-1 \).

Step 5: Horizontal Asymptote

For the function \( y=-\frac{1}{x + 1}-5 \), as \( x\to\pm\infty \), \( -\frac{1}{x + 1}\to0 \), so \( f(x)\to - 5 \). Thus, the horizontal asymptote is \( y=-5 \).

Answer:

For the function \( f(x)=\frac{1}{x - 5}+2 \):
a. Transformations: Horizontal shift right by 5 units, vertical shift up by 2 units.
b. Domain: \( (-\infty,5)\cup(5,\infty) \)
c. Range: \( (-\infty,2)\cup(2,\infty) \)
d. Vertical Asymptote: \( x = 5 \)
e. Horizontal Asymptote: \( y=2 \)

For the function \( f(x)=-\frac{1}{x + 1}-5 \):
a. Transformations: Horizontal shift left by 1 unit, reflection over the x - axis, vertical shift down by 5 units.
b. Domain: \( (-\infty,-1)\cup(-1,\infty) \)
c. Range: \( (-\infty,-5)\cup(-5,\infty) \)
d. Vertical Asymptote: \( x=-1 \)
e. Horizontal Asymptote: \( y = - 5 \)