Problem 49
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. $$ s(x)=\frac{6}{x^{2}-5 x-6} $$
Step-by-Step Solution
Verified Answer
The domain is \(x \neq 6, x \neq -1\); range is \(y \neq 0\). Vertical asymptotes: \(x = 6, x = -1\); horizontal asymptote: \(y = 0\); y-intercept: (0, -1).
1Step 1: Find the X-Intercepts
The x-intercepts are the values of \(x\) for which \(s(x) = 0\). Since \(s(x)\) is a rational function \(\left(\frac{6}{x^2 - 5x - 6}\right)\) with a constant numerator, there are no x-intercepts.
2Step 2: Find the Y-Intercept
The y-intercept is found by evaluating \(s(0)\). We substitute 0 for \(x\) in the function: \[ s(0) = \frac{6}{0^2 - 5 \times 0 - 6} = \frac{6}{-6} = -1 \]Thus, the y-intercept is \( (0, -1) \).
3Step 3: Determine Vertical Asymptotes
Vertical asymptotes occur where the denominator is zero and the function is undefined. Solve for \(x\) in the equation \(x^2 - 5x - 6 = 0\):Factoring the quadratic gives \((x - 6)(x + 1) = 0\). So, \(x = 6\) and \(x = -1\) are vertical asymptotes.
4Step 4: Determine Horizontal Asymptote
For rational functions \(f(x) = \frac{a}{b}\), the horizontal asymptote depends on the degrees of the polynomials. Here, the degree of the numerator (0) is less than the degree of the denominator (2), making the horizontal asymptote \(y = 0\).
5Step 5: State the Domain
The domain of \(s(x)\) includes all real numbers except where the denominator is zero. Therefore, the domain is all real numbers except \(x = 6\) and \(x = -1\), written as \( \{x \in \mathbb{R} \mid x eq 6, x eq -1\} \).
6Step 6: Analyze and Sketch the Graph
Using the intercepts and asymptotes, sketch the graph:- It crosses the y-axis at \( (0, -1) \).- As \(x\) approaches 6 or -1 from either side, the function approaches infinity or negative infinity. Thus, there are steep curves near these lines.- The graph will approach \(y = 0\) (horizontal asymptote) as \(x\) approaches ±∞.Use a graphing device to confirm the behavior and appearance of the function.
7Step 7: State the Range
The range of \(s(x)\) consists of all values \(y \in \mathbb{R}\) except \(y = 0\), because the graph approaches the horizontal asymptote \(y = 0\) but never reaches it. Therefore, the range is \( \{y \in \mathbb{R} \mid y eq 0\} \).
Key Concepts
InterceptsAsymptotesDomain and Range
Intercepts
Intercepts are the points where a graph crosses the axes. For rational functions like \( s(x) = \frac{6}{x^{2} - 5x - 6} \), determining intercepts is a crucial step in understanding the behavior of the graph.
**X-Intercepts**: The x-intercepts occur where \( s(x) = 0 \). However, in this function, the numerator is a constant (6), and since it cannot be zero, there are no x-intercepts. This indicates that the graph does not cross the x-axis.
**Y-Intercept**: To find the y-intercept, you substitute \( x = 0 \) into the function: \[ s(0) = \frac{6}{0^{2} - 5 \times 0 - 6} = \frac{6}{-6} = -1 \] Thus, the y-intercept is at the point \( (0, -1) \). This is where the graph crosses the y-axis.
**X-Intercepts**: The x-intercepts occur where \( s(x) = 0 \). However, in this function, the numerator is a constant (6), and since it cannot be zero, there are no x-intercepts. This indicates that the graph does not cross the x-axis.
**Y-Intercept**: To find the y-intercept, you substitute \( x = 0 \) into the function: \[ s(0) = \frac{6}{0^{2} - 5 \times 0 - 6} = \frac{6}{-6} = -1 \] Thus, the y-intercept is at the point \( (0, -1) \). This is where the graph crosses the y-axis.
Asymptotes
Asymptotes are lines that the graph of a function approaches but never actually reaches. They are essential in understanding the behavior of a rational function beyond the intercepts.
**Vertical Asymptotes**: Found by setting the denominator equal to zero and solving: \( x^2 - 5x - 6 = 0 \) This can be factored into \((x-6)(x+1) = 0\). Thus, the vertical asymptotes are at \( x = 6 \) and \( x = -1 \). The function will approach infinity or negative infinity as \( x \) gets close to these values.
**Horizontal Asymptote**: Determined by comparing the degrees of the polynomials in the numerator and denominator. Here, the degree of the numerator is 0 (since it is a constant), and the degree of the denominator is 2 (because of \( x^2 \)). When the degree of the numerator is less than the degree of the denominator, the horizontal asymptote is \( y = 0 \). This indicates that as \( x \) approaches infinity, the value of \( s(x) \) will get closer and closer to 0.
**Vertical Asymptotes**: Found by setting the denominator equal to zero and solving: \( x^2 - 5x - 6 = 0 \) This can be factored into \((x-6)(x+1) = 0\). Thus, the vertical asymptotes are at \( x = 6 \) and \( x = -1 \). The function will approach infinity or negative infinity as \( x \) gets close to these values.
**Horizontal Asymptote**: Determined by comparing the degrees of the polynomials in the numerator and denominator. Here, the degree of the numerator is 0 (since it is a constant), and the degree of the denominator is 2 (because of \( x^2 \)). When the degree of the numerator is less than the degree of the denominator, the horizontal asymptote is \( y = 0 \). This indicates that as \( x \) approaches infinity, the value of \( s(x) \) will get closer and closer to 0.
Domain and Range
The domain and range of a function inform us where the function is defined and what y-values it can take.
**Domain**: To find the domain of \( s(x) \), identify the x-values for which the function is undefined. This happens where the denominator is zero. We already found that \( x = 6 \) and \( x = -1 \) make the denominator zero, so the domain is all real numbers except at these points. This can be expressed as: \( \{ x \in \mathbb{R} \mid x eq 6, x eq -1 \} \).
**Range**: The range is the set of all possible output values (y-values) of the function. Since there is a horizontal asymptote at \( y = 0 \), the function can never actually take this value. Hence, the range of \( s(x) \) is all real numbers except y = 0. Specifically, \( \{ y \in \mathbb{R} \mid y eq 0 \} \). This means that the output of the function will be any real number except zero.
**Domain**: To find the domain of \( s(x) \), identify the x-values for which the function is undefined. This happens where the denominator is zero. We already found that \( x = 6 \) and \( x = -1 \) make the denominator zero, so the domain is all real numbers except at these points. This can be expressed as: \( \{ x \in \mathbb{R} \mid x eq 6, x eq -1 \} \).
**Range**: The range is the set of all possible output values (y-values) of the function. Since there is a horizontal asymptote at \( y = 0 \), the function can never actually take this value. Hence, the range of \( s(x) \) is all real numbers except y = 0. Specifically, \( \{ y \in \mathbb{R} \mid y eq 0 \} \). This means that the output of the function will be any real number except zero.
Other exercises in this chapter
Problem 48
Find all zeros of the polynomial. \(P(x)=x^{3}+7 x^{2}+18 x+18\)
View solution Problem 49
Find all the real zeros of the polynomial. Use the quadratic formula if necessary, as in Example \(3(\mathrm{a}) .\) $$ P(x)=x^{4}-6 x^{3}+4 x^{2}+15 x+4 $$
View solution Problem 49
\(47-50\) The graph of a polynomial function is given. From the graph, find (a) the \(x-\) and \(y\) -intercepts, and (b) the coordinates of all local extrema.
View solution Problem 49
\(39-51\) . Use synthetic division and the Remainder Theorem to evaluate \(P(c) .\) $$ P(x)=3 x^{3}+4 x^{2}-2 x+1, \quad c=\frac{2}{3} $$
View solution