Chapter 4: Problem 61
Complete the following. (a) Find the domain of \(f\) (b) Graph \(f\) in an appropriate viewing rectangle. (c) Find any horizontal or vertical asymptotes. (d) Sketch a graph of \(f\) that includes any asymptotes. $$ f(x)=\frac{x^{2}-4}{x-2} $$
Short Answer
Expert verified
Domain: all real numbers except 2. Graph: linear with a hole at x=2. No asymptotes.
Step by step solution
01
Find the Domain of f(x)
The domain of a function consists of all possible input values that do not cause division by zero or other undefined expressions. The function is given by \[ f(x) = \frac{x^2 - 4}{x - 2}. \] To find the domain, identify values of \(x\) where the denominator becomes zero. Set the denominator \(x - 2 = 0\), solving gives \(x = 2\). Therefore, the domain of \(f(x)\) is all real numbers except \(x = 2\), which can be expressed as \( (-\infty, 2) \cup (2, \infty) \).
02
Simplify f(x) if Possible
First, check for any possible simplification in the function. The numerator \(x^2 - 4\) can be factored using the difference of squares: \( (x-2)(x+2) \). Therefore, \[ f(x) = \frac{(x-2)(x+2)}{x-2}. \] This can be simplified for \(x eq 2\) to \(f(x) = x + 2\). However, at \(x = 2\), the function is undefined.
03
Graph f(x) in an Appropriate Viewing Rectangle
Using the simplified version \(y = x + 2\), draw the line \(y = x + 2\) but remember that \(f(x)\) is undefined at \(x = 2\). In the graph, include a hole at \(x = 2\) because the expression was simplified by removing a factor that causes the function to be undefined. The viewing rectangle should cover a region that clearly shows the trend of \(y = x + 2\) with an x-range around \(x = 2\).
04
Find Asymptotes
To find horizontal asymptotes, evaluate the limits of \(f(x)\) as \(x\) approaches infinity. For \(f(x) = x + 2\), there are no horizontal asymptotes as the function's slope is non-zero.Vertical asymptotes occur when the denominator equals zero. Here, \(x = 2\) is a point of discontinuity due to the removable discontinuity but it is not a vertical asymptote since \(f(x)\) is defined and linear elsewhere.
05
Sketch the Graph Including Asymptotes
Draw the line \(y = x + 2\) over the defined domain. Indicate a hole at \(x = 2\) by marking it with an open dot on the graph. Since the line \(y = x + 2\) has no horizontal or vertical asymptotes, there are no asymptotes to include in the sketch.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Domain of a function
The domain of a function refers to the set of all possible input values (typically represented by \( x \)) that will produce a valid output. In simpler terms, it's the range of \( x \)-values that you can plug into the function without breaking any mathematical rules, like dividing by zero. For the function \( f(x) = \frac{x^2 - 4}{x - 2} \), you need to find which \( x \)-values make the denominator zero because division by zero is undefined.
To find the domain of \( f(x) \):
To find the domain of \( f(x) \):
- Set the denominator equal to zero to find any restrictions: \( x - 2 = 0 \).
- Solve this equation, which gives \( x = 2 \).
- This means that \( x = 2 \) is not included in the domain.
Graphing rational functions
Graphing rational functions involves plotting the function on a coordinate plane to visualize its behavior and see where any points of discontinuity or holes occur. For the function \( f(x) = \frac{x^2 - 4}{x-2} \), you want to simplify first, if possible. Noticing the numerator can be factored as a difference of squares \((x-2)(x+2)\), the function simplifies to \( f(x) = x + 2 \) for all \( x eq 2 \).
When graphing:
When graphing:
- Draw the simplified expression \( y = x + 2 \), a straight line.
- Remember that at \( x = 2 \), the original function is undefined leading to a "hole" on the graph. Indicate this point with an open dot.
- Choose an appropriate range in the viewing rectangle, particularly surrounding \( x = 2 \), to clearly show this discontinuity.
Horizontal and vertical asymptotes
Asymptotes in a graph are lines that the graph approaches but never actually reaches. They're crucial in understanding how a function behaves at its extremes. For rational functions, horizontal asymptotes can be found by examining the behavior of the function as \( x \) approaches infinity or negative infinity, while vertical asymptotes are found where the function blows up because of division by zero.
In the case of \( f(x) = x + 2 \), which comes from simplifying \( \frac{x^2 - 4}{x-2} \):
In the case of \( f(x) = x + 2 \), which comes from simplifying \( \frac{x^2 - 4}{x-2} \):
- There is no horizontal asymptote because the simplified function \( x + 2 \) is a linear line without a horizontal limit as \( x \) increases or decreases indefinitely.
- No vertical asymptotes exist because the discontinuity at \( x = 2 \) is a removable one, evident from the factor \( (x-2) \) that cancels out.