Problem 28
Question
Sketch the graph of the rational function by hand. As sketching aids, check for intercepts, vertical asymptotes, horizontal asymptotes, and holes. Use a graphing utility to verify your graph. $$f(x)=\frac{2 x}{x^{2}+x-2}$$
Step-by-Step Solution
Verified Answer
The function \(\displaystyle f(x)=\frac{2 x}{x^{2}+x-2}\) can be simplified to \(\displaystyle f(x)=\frac{2 x}{(x-1)(x+2)}\). The only intercept is at the origin (0,0). The vertical asymptotes are at \(x = 1\) and \(x = -2\). The horizontal asymptote is at \(y = 0\). Since there are no common factors in the numerator and the denominator, the graph has no holes.
1Step 1: Simplify the Function and Identify Intercepts
First, separate the function \(\displaystyle f(x)=\frac{2 x}{x^{2}+x-2}\) into a simpler rational function by factoring its denominator. The denominator \(x^{2}+x-2 \) factors to \((x-1)(x+2)\), thus the function simplifies to \(\displaystyle f(x)=\frac{2 x}{(x-1)(x+2)}\). To get the intercepts, set \(f(x)\) to 0 and \(x\) to 0 in turn. For \(f(x) = 0\), \(x = 0\). For \(x = 0\), \(f(x) = 0\). Therefore, the only intercept is at the origin (0,0).
2Step 2: Find the Vertical Asymptotes
Vertical asymptotes are the values of \(x\) where the function tends to infinity. They happen at the points where the denominator equals 0 (because dividing by zero is undefined). Hence, we make the denominator, \((x-1)(x+2)\) equals to 0 and solve for \(x\). So, \(x = 1 \) and \(x = -2\) are our vertical asymptotes.
3Step 3: Find the Horizontal Asymptote
Horizontal asymptotes refer to the behavior of the function as \(x\) tends to infinity. Since the degree of the denominator is greater than the degree of the numerator, the function goes to 0 as \(x\) approaches to ± infinity. Thus, \(y = 0\) is our horizontal asymptote.
4Step 4: Check for Holes in the Graph
Holes in the graph occur when there is a common factor in the numerator and the denominator. However, there are no common factors between the numerator and the denominator of our function. Therefore, there are no holes in the graph of this function.
5Step 5: Sketching the Graph
Once you have analyzed all of the components, you can sketch the graph. Plot the intercepts and asymptotes first, and then draw the curves towards them. Leaving away any gaps for holes in the function. At last, confirm the drawn graph using graphing software.
Key Concepts
Vertical AsymptotesHorizontal AsymptotesGraphing Rational FunctionsIntercepts in Functions
Vertical Asymptotes
Vertical asymptotes are essential elements of rational functions. They occur where the function's denominator is zero since dividing by zero is undefined. In our function, \( f(x)=\frac{2x}{(x-1)(x+2)} \), setting the denominator equal to zero will help us find these points. We solve \((x-1)(x+2)=0\), resulting in \(x=1\) and \(x=-2\).
This means our function approaches infinity at these \(x\)-values, resulting in vertical lines known as vertical asymptotes. Remember, vertical asymptotes are not part of the graph itself but guide the behavior of the plot as it extends towards infinity. They are valuable clues in predicting how the function behaves around these points.
This means our function approaches infinity at these \(x\)-values, resulting in vertical lines known as vertical asymptotes. Remember, vertical asymptotes are not part of the graph itself but guide the behavior of the plot as it extends towards infinity. They are valuable clues in predicting how the function behaves around these points.
Horizontal Asymptotes
Horizontal asymptotes reveal how a function behaves as \(x\) approaches infinity or negative infinity. In rational functions, they occur when the degrees of the numerator and denominator influence the behavior at extreme \(x\)-values.
In \( f(x)=\frac{2x}{(x-1)(x+2)} \), the degree of the numerator is 1, and the denominator is 2. Since the denominator's degree is greater, the function approaches zero as \(x\) tends towards ± infinity. Thus, \(y = 0\) is the horizontal asymptote.
Don't forget, horizontal asymptotes guide end behavior without restricting points crossing them in the middle of the graph. The analysis of asymptotes is crucial for understanding the overall trend of the graph.
In \( f(x)=\frac{2x}{(x-1)(x+2)} \), the degree of the numerator is 1, and the denominator is 2. Since the denominator's degree is greater, the function approaches zero as \(x\) tends towards ± infinity. Thus, \(y = 0\) is the horizontal asymptote.
Don't forget, horizontal asymptotes guide end behavior without restricting points crossing them in the middle of the graph. The analysis of asymptotes is crucial for understanding the overall trend of the graph.
Graphing Rational Functions
Graphing rational functions involves understanding various components such as intercepts, asymptotes, and end behavior. Start by determining the intercepts where the function crosses the axes. In our function, the only intercept is at \((0,0)\).
Next, plot vertical asymptotes at \(x = 1\) and \(x = -2\) and draw the horizontal asymptote at \(y = 0\). Observing the sketch's behavior around these asymptotes helps in shaping the graph accurately.
Next, plot vertical asymptotes at \(x = 1\) and \(x = -2\) and draw the horizontal asymptote at \(y = 0\). Observing the sketch's behavior around these asymptotes helps in shaping the graph accurately.
- Vertical asymptotes indicate where the graph will never touch or cross.
- Horizontal asymptotes represent the leveling behavior at extreme values.
Intercepts in Functions
Intercepts are key in understanding where a graph crosses the axes. For rational functions, calculating intercepts involves setting \(x\) or \(f(x)\) to zero to find the corresponding \(y\) or \(x\)-values respectively.
In our case, setting \(x = 0\) gives \(f(x) = 0\), revealing a y-intercept at \( (0,0) \). Setting \(f(x) = 0\) also leads to \(x=0\), verifying the only intercept is at the origin.
This simple step allows identification of points where the function touches the x-axis or y-axis, forming a basis for sketching the graph. Intercepts provide concrete starting points on the graph, making them critical for the initial setup of any function's graphing task.
In our case, setting \(x = 0\) gives \(f(x) = 0\), revealing a y-intercept at \( (0,0) \). Setting \(f(x) = 0\) also leads to \(x=0\), verifying the only intercept is at the origin.
This simple step allows identification of points where the function touches the x-axis or y-axis, forming a basis for sketching the graph. Intercepts provide concrete starting points on the graph, making them critical for the initial setup of any function's graphing task.
Other exercises in this chapter
Problem 28
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