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Magazine Chapter 4: Problem 32
Sketch the graph of \(f\). $$ f(x)=\frac{x^{2}-4}{x^{2}+1} $$
Short Answer
Expert verified
The graph is symmetric about the y-axis, has x-intercepts at \((-2,0)\) and \((2,0)\), a y-intercept at \((0, -4)\), and approaches a horizontal asymptote \(y=1\).
Step by step solution
01
Identify the Domain
To find the domain of the function \( f(x) = \frac{x^2 - 4}{x^2 + 1} \), note that the denominator \( x^2 + 1 \) is never equal to zero for any real number \( x \). Therefore, the domain is all real numbers, \( x \in (-\infty, \infty) \).
02
Find the Intercepts
The function has a y-intercept when \( x = 0 \). Substitute \( x = 0 \) into the function: \( f(0) = \frac{0^2 - 4}{0^2 + 1} = -4 \). Thus, the y-intercept is at \( (0, -4) \). Next, set the numerator equal to zero to find x-intercepts: \( x^2 - 4 = 0 \) which gives \( x = \pm 2 \). Thus, the x-intercepts are \( (2, 0) \) and \( (-2, 0) \).
03
Analyze Behavior at Infinity
Consider the limits as \( x \to \infty \) and \( x \to -\infty \): \( \lim_{x \to \pm \infty} \frac{x^2 - 4}{x^2 + 1} = 1 \). The graph approaches the horizontal asymptote \( y = 1 \) as \( x \to \pm\infty \).
04
Determine the Symmetry of the Function
Check for symmetry by substituting \( -x \) into the function: \( f(-x) = \frac{(-x)^2 - 4}{(-x)^2 + 1} = \frac{x^2 - 4}{x^2 + 1} = f(x) \). This confirms that the function is even, so the graph is symmetric about the y-axis.
05
Analyze the Behavior Around Critical Points
Find the derivative \( f'(x) = \frac{(2x)(x^2+1) - (x^2-4)(2x)}{(x^2+1)^2} = \frac{-8x}{(x^2+1)^2} \). The critical point is at \( x = 0 \), where \( f'(x) = 0 \). This point \( (0, -4) \) determines a local minimum since \( f'(x) < 0 \) for \( x > 0 \) and \( f'(x) > 0 \) for \( x < 0 \). Also, \( f \) is decreasing on \( x > 0 \) and increasing on \( x < 0 \).
06
Sketch the Graph
Using the information from previous steps, sketch the graph. Begin with plotting intercepts \((0, -4)\), \((2,0)\), and \((-2,0)\). Draw the horizontal asymptote \( y = 1 \). The function is symmetric about the y-axis. The graph decreases from infinity towards a local minimum at \( (0, -4) \) and then increases back to infinity, approaching \( y = 1 \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Domain of a Function
When graphing a rational function such as \( f(x) = \frac{x^2 - 4}{x^2 + 1} \), understanding the domain is crucial. The domain refers to all the possible input values (or \( x \)-values) that won't make the function undefined.
In this function, the denominator \( x^2 + 1 \) cannot equal zero. Since \( x^2 + 1 > 0 \) for all real numbers, this function is never undefined. Therefore, its domain includes all real numbers: \( x \in (-\infty, \infty) \).
Recognizing the domain helps us know where we can and cannot evaluate the function, ensuring our graph remains accurate and complete.
In this function, the denominator \( x^2 + 1 \) cannot equal zero. Since \( x^2 + 1 > 0 \) for all real numbers, this function is never undefined. Therefore, its domain includes all real numbers: \( x \in (-\infty, \infty) \).
Recognizing the domain helps us know where we can and cannot evaluate the function, ensuring our graph remains accurate and complete.
X-Intercepts and Y-Intercepts
Identifying intercepts on a graph offers valuable insight into where the graph crosses the axes. The x-intercepts occur where the graph intersects the x-axis, which happens when the function equals zero:
For \( f(x) = \frac{x^2 - 4}{x^2 + 1} \), set the numerator equal to zero: \( x^2 - 4 = 0 \). Solving this gives \( x = \pm 2 \). Thus, the x-intercepts are \((2, 0)\) and \((-2, 0)\).
The y-intercept occurs at the point where the graph crosses the y-axis. This is when \( x = 0 \). Substituting \( x = 0 \) into the function: \( f(0) = \frac{0^2 - 4}{0^2 + 1} = -4 \). Hence, the y-intercept is at \((0, -4)\).
Plotting these intercepts is essential for accurately sketching the graph.
For \( f(x) = \frac{x^2 - 4}{x^2 + 1} \), set the numerator equal to zero: \( x^2 - 4 = 0 \). Solving this gives \( x = \pm 2 \). Thus, the x-intercepts are \((2, 0)\) and \((-2, 0)\).
The y-intercept occurs at the point where the graph crosses the y-axis. This is when \( x = 0 \). Substituting \( x = 0 \) into the function: \( f(0) = \frac{0^2 - 4}{0^2 + 1} = -4 \). Hence, the y-intercept is at \((0, -4)\).
Plotting these intercepts is essential for accurately sketching the graph.
Horizontal Asymptote
Asymptotes are lines that the graph approaches but never touches. For a rational function, horizontal asymptotes tell us the behavior of the graph as \( x \) becomes very large (positive or negative).
For \( f(x) = \frac{x^2 - 4}{x^2 + 1} \), analyze the limit as \( x \to \pm\infty \). Replace all terms in the function with their leading terms: \( \frac{x^2}{x^2} = 1 \). Therefore, \( \lim_{x \to \pm\infty} f(x) = 1 \), indicating a horizontal asymptote at \( y = 1 \).
Understanding this asymptotic behavior helps define the edges of the graph and visualize its long-term trend.
For \( f(x) = \frac{x^2 - 4}{x^2 + 1} \), analyze the limit as \( x \to \pm\infty \). Replace all terms in the function with their leading terms: \( \frac{x^2}{x^2} = 1 \). Therefore, \( \lim_{x \to \pm\infty} f(x) = 1 \), indicating a horizontal asymptote at \( y = 1 \).
Understanding this asymptotic behavior helps define the edges of the graph and visualize its long-term trend.
Symmetry of Functions
Recognizing symmetry makes sketching rational functions easier and helps anticipate certain graph behaviors. A function is even if it is symmetric about the y-axis. Mathematically, this means \( f(-x) = f(x) \) for all \( x \).
Applying this to \( f(x) = \frac{x^2 - 4}{x^2 + 1} \), check: \( f(-x) = \frac{(-x)^2 - 4}{(-x)^2 + 1} = \frac{x^2 - 4}{x^2 + 1} = f(x) \). Thus, the function is indeed even.
This symmetry simplifies plotting as once you know one side of the y-axis, you automatically know the other. Moreover, it reinforces the correct placement of intercepts and asymptotical behavior discussed earlier.
Applying this to \( f(x) = \frac{x^2 - 4}{x^2 + 1} \), check: \( f(-x) = \frac{(-x)^2 - 4}{(-x)^2 + 1} = \frac{x^2 - 4}{x^2 + 1} = f(x) \). Thus, the function is indeed even.
This symmetry simplifies plotting as once you know one side of the y-axis, you automatically know the other. Moreover, it reinforces the correct placement of intercepts and asymptotical behavior discussed earlier.
