Problem 55
Question
Find the intercepts and asymptotes, and then sketch a graph of the rational function and state the domain and range. Use a graphing device to confirm your answer. $$ r(x)=\frac{x^{2}-2 x+1}{x^{2}+2 x+1} $$
Step-by-Step Solution
Verified Answer
Domain: All real numbers except \(x = -1\). Range: All real numbers except \(y = 1\).
1Step 1: Factor the Rational Function
First, we will factor the numerator and the denominator of the rational function. The numerator is \(x^2 - 2x + 1\), which can be factored as \((x - 1)^2\). The denominator is \(x^2 + 2x + 1\), which can be factored as \((x + 1)^2\). Thus, the function becomes: \[ r(x) = \frac{(x - 1)^2}{(x + 1)^2} \]
2Step 2: Find the Intercepts
To find the x-intercept, set the numerator equal to zero: \((x - 1)^2 = 0\), which gives \(x = 1\). To find the y-intercept, set \(x = 0\): \[ r(0) = \frac{(0 - 1)^2}{(0 + 1)^2} = 1 \]. So the intercepts are (1, 0) for the x-intercept and (0, 1) for the y-intercept.
3Step 3: Determine Vertical Asymptotes
Vertical asymptotes occur where the denominator equals zero but the numerator does not. However, both the numerator and the denominator become zero at \(x = -1\), resulting in a hole rather than an asymptote at this point.
4Step 4: Determine Horizontal Asymptotes
Since the degrees of the numerator and denominator are equal, the horizontal asymptote is determined by the ratio of the leading coefficients. Hence, the horizontal asymptote is \(y = 1\).
5Step 5: Sketch the Graph
The graph has a hole at \(x = -1\) and approaches the horizontal asymptote \(y = 1\). It passes through the intercepts at (1, 0) and (0, 1). The sketch will show these features.
6Step 6: Determine the Domain and Range
The domain excludes the x-value where the hole occurs, so \(x eq -1\). Therefore, the domain is all real numbers except \(-1\): \(x \in \mathbb{R}, x eq -1\). The range is all real numbers except where the horizontal asymptote is (since the function never actually reaches it): \(y eq 1\).
Key Concepts
InterceptsAsymptotesDomain and Range
Intercepts
When graphing a rational function, finding the intercepts is the first essential step. An intercept is a point where the graph crosses an axis. There are two types of intercepts to consider: the x-intercept and the y-intercept.
To find the **x-intercept**, we need to set the numerator of the rational function equal to zero because at the x-intercept, the function value is zero, meaning the whole fraction becomes zero. For the function \[ r(x) = \frac{(x - 1)^2}{(x + 1)^2}\]we'll set \( (x - 1)^2 = 0 \), which yields \( x = 1 \). Therefore, the x-intercept is at the point (1, 0).
The **y-intercept** is found by substituting \( x = 0 \) into the function, as this will give the point where the graph crosses the y-axis. In our example:\[ r(0) = \frac{(0 - 1)^2}{(0 + 1)^2} = 1 \]This means the y-intercept is at (0, 1). Identifying intercepts helps in sketching the initial framework of the graph.
To find the **x-intercept**, we need to set the numerator of the rational function equal to zero because at the x-intercept, the function value is zero, meaning the whole fraction becomes zero. For the function \[ r(x) = \frac{(x - 1)^2}{(x + 1)^2}\]we'll set \( (x - 1)^2 = 0 \), which yields \( x = 1 \). Therefore, the x-intercept is at the point (1, 0).
The **y-intercept** is found by substituting \( x = 0 \) into the function, as this will give the point where the graph crosses the y-axis. In our example:\[ r(0) = \frac{(0 - 1)^2}{(0 + 1)^2} = 1 \]This means the y-intercept is at (0, 1). Identifying intercepts helps in sketching the initial framework of the graph.
Asymptotes
Asymptotes are lines that the graph of the function approaches but never actually reaches. They help determine the end behavior of the rational function and provide insight into what happens as the x-values become very large or very small. There are typically two types considered in rational functions: vertical and horizontal asymptotes.
For **vertical asymptotes**, we look at the values that make the denominator zero; however, they only become vertical asymptotes if the numerator is non-zero at those same values. In the function \[r(x) = \frac{(x - 1)^2}{(x + 1)^2}\]when \(x = -1\), both the numerator and the denominator become zero, indicating a hole instead of an asymptote at \(x = -1\).
**Horizontal asymptotes** are determined based on the degrees of the polynomials in the numerator and the denominator. Here, both are of degree 2. So, we take the ratio of their leading coefficients, which is 1, resulting in a horizontal asymptote at \(y = 1\). This asymptote tells us about the behavior of the function as \(x\) approaches infinity or negative infinity.
For **vertical asymptotes**, we look at the values that make the denominator zero; however, they only become vertical asymptotes if the numerator is non-zero at those same values. In the function \[r(x) = \frac{(x - 1)^2}{(x + 1)^2}\]when \(x = -1\), both the numerator and the denominator become zero, indicating a hole instead of an asymptote at \(x = -1\).
**Horizontal asymptotes** are determined based on the degrees of the polynomials in the numerator and the denominator. Here, both are of degree 2. So, we take the ratio of their leading coefficients, which is 1, resulting in a horizontal asymptote at \(y = 1\). This asymptote tells us about the behavior of the function as \(x\) approaches infinity or negative infinity.
Domain and Range
The domain of a rational function is the set of all possible x-values that can be plugged into the function without making the denominator zero. For our function \[ r(x) = \frac{(x - 1)^2}{(x + 1)^2}\]we identified earlier that \(x = -1\) makes the function undefined due to a hole rather than an asymptote. Hence, the domain excludes \(x = -1\), expressed as all real numbers except -1: \( x \in \mathbb{R}, x eq -1 \).
The range refers to all possible y-values the function can take. Considering the horizontal asymptote at \(y = 1\) and knowing that the function doesn’t actually reach this line, the range excludes \(y = 1\). It includes all other real numbers, stated as: \( y \in \mathbb{R}, y eq 1 \).
Understanding the domain and range is crucial to fully characterize the behavior of the rational function and accurately represent it graphically.
The range refers to all possible y-values the function can take. Considering the horizontal asymptote at \(y = 1\) and knowing that the function doesn’t actually reach this line, the range excludes \(y = 1\). It includes all other real numbers, stated as: \( y \in \mathbb{R}, y eq 1 \).
Understanding the domain and range is crucial to fully characterize the behavior of the rational function and accurately represent it graphically.
Other exercises in this chapter
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