Problem 32
Question
Sketch the graph of \(f\). $$ f(x)=\frac{x^{2}-4}{x^{2}+1} $$
Step-by-Step Solution
Verified Answer
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 \(y=1\).
1Step 1: 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, \( x \in (-\infty, \infty) \).
2Step 2: 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 \( (-2, 0) \).
3Step 3: 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 \( x \to \pm\infty \).
4Step 4: 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.
5Step 5: 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 \( x < 0 \).
6Step 6: 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 \( y = 1 \).
Key Concepts
Domain of a FunctionX-Intercepts and Y-InterceptsHorizontal AsymptoteSymmetry of Functions
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.
Other exercises in this chapter
Problem 31
Sketch the graph of \(f\). $$ f(x)=\frac{-3 x^{2}}{x^{2}+1} $$
View solution Problem 32
Examine the expression for the given set of data points of the form \((x, y)\). Find the constant of variation and a formula that describes how \(y\) varies wit
View solution Problem 32
The frame for a shipping crate is to be constructed from 24 feet of \(2 \times 2\) lumber. Assuming the crate is to have square ends of length \(x\) feet, deter
View solution Problem 33
Examine the expression for the given set of data points of the form \((x, y)\). Find the constant of variation and a formula that describes how \(y\) varies wit
View solution