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Magazine Chapter 3: Problem 61
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. $$r(x)=\frac{5 x^{2}+5}{x^{2}+4 x+4}$$
Short Answer
Expert verified
Y-intercept: \(\frac{5}{4}\), no x-intercepts; asymptotes: vertical at \(x = -2\) and horizontal at \(y = 5\); domain: \((-\infty, -2) \cup (-2, \infty)\), range: \((-\infty, 5) \cup (5, \infty)\).
Step by step solution
01
Find the Y-intercept
To find the y-intercept, set \( x = 0 \) in the function \( r(x) = \frac{5x^2 + 5}{x^2 + 4x + 4} \). This gives us \( r(0) = \frac{5(0)^2 + 5}{(0)^2 + 4(0) + 4} = \frac{5}{4} \). Therefore, the y-intercept is \( \frac{5}{4} \).
02
Find the X-intercepts
To find the x-intercepts, set the numerator of the rational function equal to zero: \( 5x^2 + 5 = 0 \). Solving for \( x \) gives \( x^2 = -1 \), which has no real solutions. Therefore, there are no x-intercepts for \( r(x) \).
03
Identify Vertical Asymptotes
Vertical asymptotes occur where the denominator is zero and the numerator is not zero. Set the denominator equal to zero: \( x^2 + 4x + 4 = 0 \). Factoring gives \( (x+2)^2 = 0 \), so \( x = -2 \). Thus, the vertical asymptote is at \( x = -2 \).
04
Identify Horizontal Asymptotes
The horizontal asymptote can be found by comparing the degrees of the polynomial in the numerator and denominator. Both are degree 2 polynomials. Therefore, the horizontal asymptote is determined by the ratio of the leading coefficients, which is \( \frac{5}{1} = 5 \). Thus, the horizontal asymptote is \( y = 5 \).
05
Determine the Domain
The domain of the function \( r(x) \) includes all real numbers except where the denominator is zero. Since \( x = -2 \) makes the denominator zero, the domain is \( x \in (-\infty, -2) \cup (-2, \infty) \).
06
Analyze the Range
To determine the range, consider the behavior of the function and the horizontal asymptote. As \( x \) approaches infinity, \( r(x) \rightarrow 5 \), but never actually equals \( 5 \) because of the horizontal asymptote. The range is \( y \in (-\infty, 5) \cup (5, \infty) \).
07
Sketch the Graph
Using the information obtained, we can sketch the graph of \( r(x) \). The graph does not cross the x-axis since there are no x-intercepts, has a vertical asymptote at \( x = -2 \), and a horizontal asymptote at \( y = 5 \). Plot the y-intercept at \( y = \frac{5}{4} \) and consider symmetry or transformations to sketch. Verify with a graphing device to confirm the shape.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Intercepts
Intercepts are points where the graph of a function crosses the axes. For rational functions, we usually seek both the y-intercept and any x-intercepts.
- Y-intercept: This is where the graph crosses the y-axis, occurring when the input value, \( x \), is zero. In our function \( r(x) = \frac{5x^2 + 5}{x^2 + 4x + 4} \), we find the y-intercept by setting \( x = 0 \). Substituting this into the function gives \( r(0) = \frac{5}{4} \). So, the y-intercept is at \( (0, \frac{5}{4}) \).
- X-intercepts: These occur where the function crosses the x-axis, which implies \( r(x) = 0 \). For this to happen, the numerator must be zero while the denominator is non-zero. Setting \( 5x^2 + 5 = 0 \) leads to \( x^2 = -1 \). This equation has no real solutions, as the square of a real number cannot be negative. Therefore, the function has no x-intercepts.
Asymptotes
Asymptotes are lines that the graph of a function approaches but never actually reaches. They can be vertical, horizontal, or slant. For our rational function, both vertical and horizontal asymptotes are significant.
- Vertical Asymptotes: These occur where the denominator equals zero, creating a division by zero, while the numerator does not equal zero. In \( r(x) \), the denominator \( x^2 + 4x + 4 = 0 \) factors to \( (x+2)^2 = 0 \), giving a vertical asymptote at \( x = -2 \).
- Horizontal Asymptotes: Found by comparing the degrees of the numerator and the denominator. Here, both have degree 2. The horizontal asymptote is the ratio of the leading coefficients, \( \frac{5}{1} = 5 \). Thus, the horizontal asymptote is \( y = 5 \).
Domain and Range
Understanding the domain and range of a rational function helps to identify all possible input (x-values) and output (y-values).
- Domain: This refers to all the input values \( x \) for which the function is defined. A rational function is undefined where its denominator is zero. Since the denominator \( x^2 + 4x + 4 \) is zero at \( x = -2 \), the domain is all real numbers except \( x = -2 \). This can be expressed as \( x \in (-\infty, -2) \cup (-2, \infty) \).
- Range: The range is the set of all possible output values \( y \). Considering the horizontal asymptote at \( y = 5 \), the outputs can come as close to 5 as desired but will never actually reach it. Therefore, the range is \( y \in (-\infty, 5) \cup (5, \infty) \).
Graphing Equations
Graphing a rational function involves plotting important features such as intercepts and asymptotes, and understanding its overall behavior.
- Start by plotting the y-intercept at \( (0, \frac{5}{4}) \).
- Note the absence of x-intercepts, meaning the graph never touches the x-axis.
- Draw the vertical asymptote as a dashed line at \( x = -2 \). The graph cannot cross or touch this line.
- Similarly, sketch the horizontal asymptote at \( y = 5 \) as a dash, indicating the graph approaches this line infinitely but never reaches it.
- With the intercepts and asymptotes plotted, sketch the overall shape of the graph. It will curve around these guides without crossing the asymptotes, increasing or decreasing without bound.
