Chapter 3: Problem 52
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. $$s(x)=\frac{x+2}{(x+3)(x-1)}$$
Short Answer
Expert verified
Intercepts: x-intercept at (-2, 0), y-intercept at (0, -2/3). Asymptotes: vertical at x = -3 and x = 1, horizontal at y = 0. Domain: all reals except x = -3 and x = 1.
Step by step solution
01
Finding the x-intercept
To find the x-intercepts of the function \(s(x) = \frac{x+2}{(x+3)(x-1)}\), set the numerator equal to zero. Thus, solve \(x + 2 = 0\), giving \(x = -2\). So, the x-intercept is at \((-2, 0)\).
02
Finding the y-intercept
The y-intercept is found by substituting \(x = 0\) into the function. Therefore, \(s(0) = \frac{0+2}{(0+3)(0-1)} = \frac{2}{-3} = -\frac{2}{3}\). Thus, the y-intercept is at \(\left(0, -\frac{2}{3}\right)\).
03
Finding the vertical asymptotes
Vertical asymptotes occur where the denominator is zero, provided the numerator is not zero at those points. For \(s(x)\), the denominator is \((x+3)(x-1)\). Setting this equal to zero, solve \((x+3) = 0\) and \((x-1) = 0\). The solutions are \(x = -3\) and \(x = 1\); thus vertical asymptotes are at \(x = -3\) and \(x = 1\).
04
Finding the horizontal asymptote
For a rational function \(\frac{ax^m}{bx^n}\), where \(m\) and \(n\) are the degrees of the polynomials in the numerator and denominator, the horizontal asymptote is determined as follows: if the degree of the denominator is greater than the numerator, the horizontal asymptote is \(y = 0\). In this case, the numerator is degree 1, and the denominator is degree 2, so the horizontal asymptote is \(y = 0\).
05
Determining the domain of the function
The domain of \(s(x)\) excludes the x-values that make the denominator zero. Thus, the function is undefined at \(x = -3\) and \(x = 1\). Therefore, the domain is all real numbers except \(x = -3\) and \(x = 1\), or \((-\infty, -3) \cup (-3, 1) \cup (1, \infty)\).
06
Sketching the graph and confirming with a device
Start by plotting the x-intercept \((-2, 0)\) and the y-intercept \(\left(0, -\frac{2}{3}\right)\). Draw dashed lines for the vertical asymptotes at \(x = -3\) and \(x = 1\), and the horizontal asymptote at \(y = 0\). Then sketch the graph approaching these asymptotes and intercepts. Use a graphing device to check the shape and features of \(s(x)\) for accuracy.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Intercepts
When working with rational functions like \(s(x) = \frac{x+2}{(x+3)(x-1)}\), finding intercepts can help you understand where the graph of the function will cross the axes.
To find the **x-intercepts**, set the numerator equal to zero and solve for \(x\). For our function, this means solving \(x + 2 = 0\), resulting in \(x = -2\). Hence, the x-intercept is at the point \((-2, 0)\). This is where the graph touches the x-axis.
The **y-intercept** is found by setting \(x = 0\) in the function and solving for \(s(x)\). Substituting zero gives \(s(0) = \frac{2}{-3}\), leading to the y-intercept at \((0, -\frac{2}{3})\). This point indicates where the graph crosses the y-axis. Spotting these intercepts is fundamental, as they provide essential markers on which the rest of your graph will build.
To find the **x-intercepts**, set the numerator equal to zero and solve for \(x\). For our function, this means solving \(x + 2 = 0\), resulting in \(x = -2\). Hence, the x-intercept is at the point \((-2, 0)\). This is where the graph touches the x-axis.
The **y-intercept** is found by setting \(x = 0\) in the function and solving for \(s(x)\). Substituting zero gives \(s(0) = \frac{2}{-3}\), leading to the y-intercept at \((0, -\frac{2}{3})\). This point indicates where the graph crosses the y-axis. Spotting these intercepts is fundamental, as they provide essential markers on which the rest of your graph will build.
Asymptotes
Asymptotes are lines that the graph of a rational function approaches but never quite touches.
Our rational function has two types of asymptotes: **vertical** and **horizontal**. **Vertical asymptotes** occur where the denominator of the function equals zero, provided the numerator isn't zero. For the function \(s(x)\), the denominator is \((x+3)(x-1)\). By setting this equal to zero, we find \(x = -3\) and \(x = 1\) as the points where vertical asymptotes occur. These lines vertically cross the x-axis and the graph will get infinitely close to them, but not touch.
Now considering the **horizontal asymptote**, for rational functions where the degree of the numerator is less than that of the denominator, the horizontal asymptote is \(y = 0\). In our function, the numerator has degree 1 while the denominator has degree 2. Therefore, the horizontal asymptote is at \(y = 0\). This line signifies the value the function will web towards but never hit as it extends towards infinity on either side.
Our rational function has two types of asymptotes: **vertical** and **horizontal**. **Vertical asymptotes** occur where the denominator of the function equals zero, provided the numerator isn't zero. For the function \(s(x)\), the denominator is \((x+3)(x-1)\). By setting this equal to zero, we find \(x = -3\) and \(x = 1\) as the points where vertical asymptotes occur. These lines vertically cross the x-axis and the graph will get infinitely close to them, but not touch.
Now considering the **horizontal asymptote**, for rational functions where the degree of the numerator is less than that of the denominator, the horizontal asymptote is \(y = 0\). In our function, the numerator has degree 1 while the denominator has degree 2. Therefore, the horizontal asymptote is at \(y = 0\). This line signifies the value the function will web towards but never hit as it extends towards infinity on either side.
Domain and Range
The **domain** and **range** of a rational function describe the set of possible inputs (x-values) and outputs (y-values), respectively.
For a rational function such as \(s(x) = \frac{x+2}{(x+3)(x-1)}\), the domain is determined by where the function is undefined—typically where the denominator is zero. Here, \(x = -3\) and \(x = 1\) are such points. As these values make the function undefined, the domain consists of all real numbers except \(-3\) and \(1\). It is expressed in interval notation as \((-abla, -3) \cup (-3, 1) \cup (1, abla)\).
The **range** of the function involves understanding what y-values the function can take. With the horizontal asymptote at \(y = 0\), and knowing vertical asymptotes generally slice the y-values being achieved by the function, the graph of \(s(x)\) can approach but not include \(y = 0\). Similarly, the function can attain nearly all y-values except the horizontal asymptote, so the range is all real numbers except \(0\), written as \((-abla, 0) \cup (0, abla)\). Understanding these helps in visualizing how the function's values behave over intervals.
For a rational function such as \(s(x) = \frac{x+2}{(x+3)(x-1)}\), the domain is determined by where the function is undefined—typically where the denominator is zero. Here, \(x = -3\) and \(x = 1\) are such points. As these values make the function undefined, the domain consists of all real numbers except \(-3\) and \(1\). It is expressed in interval notation as \((-abla, -3) \cup (-3, 1) \cup (1, abla)\).
The **range** of the function involves understanding what y-values the function can take. With the horizontal asymptote at \(y = 0\), and knowing vertical asymptotes generally slice the y-values being achieved by the function, the graph of \(s(x)\) can approach but not include \(y = 0\). Similarly, the function can attain nearly all y-values except the horizontal asymptote, so the range is all real numbers except \(0\), written as \((-abla, 0) \cup (0, abla)\). Understanding these helps in visualizing how the function's values behave over intervals.