Question

7.6: graphing rational functions practic
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identify the holes, vertical asymptotes, x-intercepts,
sketch the graph.
(1) $f(x)=\frac{4}{x - 3}$
va:
ha:
x - int
domain:
(3) $f(x)=\frac{1}{-x + 4}$
va:
ha:
x - int:
domain:

Explanation:

Problem 1: \( f(x) = \frac{4}{x - 3} \)

Step 1: Find Vertical Asymptote (VA)

Vertical asymptotes occur where the denominator is zero (and numerator is non - zero). For \( f(x)=\frac{4}{x - 3} \), set \( x-3 = 0 \).
\( x-3=0\Rightarrow x = 3 \)

Step 2: Find Horizontal Asymptote (HA)

For a rational function \( f(x)=\frac{N(x)}{D(x)} \), if the degree of \( N(x) \) (degree of numerator, here degree of 4 is 0) is less than the degree of \( D(x) \) (degree of \( x - 3 \) is 1), the horizontal asymptote is \( y = 0 \).

Step 3: Find x - intercept

To find x - intercept, set \( f(x)=0 \), i.e., \( \frac{4}{x - 3}=0 \). But \( \frac{4}{x - 3}=0 \) has no solution since the numerator is 4 (non - zero). So, no x - intercept.

Step 4: Find Domain

The domain is all real numbers except where the denominator is zero. So, domain is \( \{x\in\mathbb{R}|x
eq3\} \) or \( (-\infty,3)\cup(3,\infty) \)

Step 1: Find Vertical Asymptote (VA)

Set the denominator \( -x + 4=0 \). Solving for x: \( -x+4 = 0\Rightarrow x = 4 \)

Step 2: Find Horizontal Asymptote (HA)

The degree of the numerator (degree of 1 is 0) is less than the degree of the denominator (degree of \( -x + 4 \) is 1). So, horizontal asymptote is \( y = 0 \)

Step 3: Find x - intercept

Set \( f(x)=0 \), i.e., \( \frac{1}{-x + 4}=0 \). Since the numerator is 1 (non - zero), there is no solution. So, no x - intercept.

Step 4: Find Domain

The domain is all real numbers except where the denominator is zero. So, domain is \( \{x\in\mathbb{R}|x
eq4\} \) or \( (-\infty,4)\cup(4,\infty) \)

Answer:

• VA: \( x = 3 \)
• HA: \( y = 0 \)
• x - int: None
• Domain: \( (-\infty,3)\cup(3,\infty) \)

Problem 3: \( f(x)=\frac{1}{-x + 4}=\frac{1}{-(x - 4)}=\frac{1}{x - 4}\times(- 1) \)