Problem 31
Question
Graph each of the functions. $$f(x)=\frac{2}{x-1}+3$$
Step-by-Step Solution
Verified Answer
Graph the rational function with asymptotes at \(x=1\) and \(y=3\).
1Step 1: Identify Key Features
The given function is \( f(x) = \frac{2}{x-1} + 3 \). This is a rational function. The function has a vertical asymptote at \( x = 1 \) since the denominator becomes zero at \( x = 1 \). The horizontal asymptote is \( y = 3 \), indicated by the constant term in the equation.
2Step 2: Determine Domain and Range
The domain of the function includes all real numbers except where the denominator is zero, which is \( x = 1 \). Hence, the domain is \( x eq 1 \). The range is all real numbers except \( y = 3 \), due to the horizontal asymptote.
3Step 3: Calculate Intercepts
First, find the y-intercept by setting \( x = 0 \): \( f(0) = \frac{2}{0-1} + 3 = -2 + 3 = 1 \). Thus, the y-intercept is \( (0, 1) \). Finding the x-intercept involves solving \( \frac{2}{x-1} + 3 = 0 \). Solving gives \( x = 1.66667 \), so the x-intercept is around \( (1.67, 0) \).
4Step 4: Create a Graph Outline
Plot the vertical asymptote (dashed line at \( x = 1 \)) and the horizontal asymptote (dashed line at \( y = 3 \)). Mark the intercepts \( (0, 1) \) and \( (1.67, 0) \).
5Step 5: Plot Additional Points
To get a better idea of the curve, test additional values like \( x = 2 \), yielding \( f(2) = \frac{2}{2-1} + 3 = 5 \), and \( x = 3 \), yielding \( f(3) = \frac{2}{3-1} + 3 = 4 \). Plot these points on the graph.
6Step 6: Sketch the Curve
Using the plotted points and asymptotes, sketch the rational function. It should approach the vertical asymptote \( x = 1 \) from both sides and extend towards \( y = 3 \) as \( x \) moves positively or negatively away from the asymptote.
Key Concepts
Vertical AsymptotesHorizontal AsymptotesDomain and Range
Vertical Asymptotes
Vertical Asymptotes are imaginary vertical lines that the graph of a function approaches but never actually touches or crosses. They occur at the values of \( x \) where the denominator of a rational function is zero. In the function \( f(x) = \frac{2}{x-1} + 3 \), the vertical asymptote is located where \( x-1 = 0 \).
This means the vertical asymptote is at \( x = 1 \). This line indicates that the function values grow infinitely larger or smaller as \( x \) approaches 1 from either direction.
This means the vertical asymptote is at \( x = 1 \). This line indicates that the function values grow infinitely larger or smaller as \( x \) approaches 1 from either direction.
- The graph will dramatically rise or fall near this line.
- The function is undefined at this point, hence the graph cannot touch or cross it.
Horizontal Asymptotes
Horizontal Asymptotes describe the behavior of a function as \( x \) goes to infinity (either positive or negative infinity). They give a sense of 'limit' that the function approaches but never actually reaches. For the function \( f(x) = \frac{2}{x-1} + 3 \), the horizontal asymptote can be identified by looking at the constant term that represents a shift. In this case, it's \( y = 3 \).
This indicates that as you move along the graph far to the left or right, the function's value will get closer and closer to 3, but it will never actually reach this value.
This indicates that as you move along the graph far to the left or right, the function's value will get closer and closer to 3, but it will never actually reach this value.
- The horizontal asymptote helps you understand the long-term behavior of the function.
- It represents the y-value that the function's output approaches as \( x \) becomes very large or very small.
Domain and Range
The Domain and Range of a function give you an overview of the values that the function can accept for \( x \) and output for \( y \). The domain consists of all the permissible inputs, while the range is the potential outputs.
For the function \( f(x) = \frac{2}{x-1} + 3 \):
For the function \( f(x) = \frac{2}{x-1} + 3 \):
- Domain: Excludes the point where the denominator is zero. Here, it means \( x = 1 \) is not included, so the domain is "all real numbers except \( x = 1 \)".
- Range: Determined by the role of the horizontal asymptote. This function can take any value except \( y = 3 \). Thus, the range is "all real numbers except \( y = 3 \)".
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