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Magazine Chapter 3: Problem 32
Sketch a graph of the function. $$ f(x)=\frac{x}{x+2} $$
Short Answer
Expert verified
Vertical asymptote at \(x = -2\); horizontal asymptote at \(y = 1\). x- and y-intercept at (0,0).
Step by step solution
01
- Identify Asymptotes
Determine the vertical and horizontal asymptotes of the function. The vertical asymptote occurs where the denominator is zero: \[ x + 2 = 0 \Rightarrow x = -2 \] For the horizontal asymptote, as \( x \) approaches infinity, the function approaches: \[ \lim_{{x \to \infty}} \frac{x}{x+2} = 1 \] So, the horizontal asymptote is \( y = 1 \).
02
- Calculate Intercepts
Find the x-intercept and y-intercept. For the x-intercept, set \( f(x) = 0 \): \[ \frac{x}{x+2} = 0 \Rightarrow x = 0 \] So the x-intercept is \( (0,0) \). For the y-intercept, substitute \( x = 0 \): \[ f(0) = \frac{0}{0+2} = 0 \] So the y-intercept is \( (0,0) \).
03
- Analyze Behavior Near Asymptotes
Examine the function's behavior as it approaches the vertical asymptote at \( x = -2 \). As \( x \to -2^+ \) and \( x \to -2^- \), the function will tend to \( \infty \) and \( -\infty \) respectively.
04
- Create a Table of Values
Choose a few values of \( x \) to calculate corresponding values of \( f(x) \). For example: \[ f(-4) = \frac{-4}{-4+2} = 2 \] \[ f(-1) = \frac{-1}{-1+2} = -1 \] \[ f(1) = \frac{1}{1+2} \approx 0.33 \] Use these points to plot on the graph.
05
- Sketch the Graph
Draw the asymptotes as dashed lines at \( x = -2 \) (vertical asymptote) and \( y = 1 \) (horizontal asymptote). Plot the intercepts and the points from the table. Sketch the curve approaching the asymptotes and passing through the plotted points.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Vertical Asymptote
A vertical asymptote represents a value of x where the function approaches infinity or negative infinity. To find the vertical asymptote for the function \( f(x)=\frac{x}{x+2} \), we focus on the denominator. We set it to zero: \( x + 2 = 0 \). From this equation, we get \( x = -2 \). This means that as \( x \) gets closer to -2 from either side, the function value becomes very large (positively or negatively), creating a vertical asymptote at \( x = -2 \). Make sure to draw this as a dashed line on your graph.
Horizontal Asymptote
A horizontal asymptote represents the value that the function approaches as \( x \) goes to infinity or negative infinity. For \( f(x)=\frac{x}{x+2} \), as \( x \to \infty \), the effect of the \( +2 \) in the denominator becomes negligible. So, the function simplifies to \( \frac{x}{x} = 1 \). Therefore, the horizontal asymptote is \( y = 1 \). You can draw this as another dashed line on your graph.
Intercepts
Finding intercepts is crucial for sketching graphs accurately. To determine the x-intercept, set \( f(x) \) to 0 and solve for \( x \): \( \frac{x}{x+2} = 0 \Rightarrow x = 0 \). Hence, the x-intercept is (0, 0). To find the y-intercept, substitute \( x = 0 \) into the function: \( f(0) = \frac{0}{0+2} = 0 \). Thus, the y-intercept is also (0, 0). These intercepts tell us where the graph crosses the x-axis and y-axis.
Behavior Near Asymptotes
Understanding how the function behaves near asymptotes helps in drawing a more precise graph. For \( f(x)=\frac{x}{x+2} \), observe the behavior as \( x \) approaches the vertical asymptote at \( x = -2 \). As \( x \to -2^+ \), \( f(x) \) tends to infinity. Conversely, as \( x \to -2^- \), \( f(x) \) tends to negative infinity. This behavior shows how the graph will dramatically rise or fall as it nears the vertical asymptote.
Table of Values
A table of values helps in plotting specific points on the graph. For instance: If \( x = -4 \): \( f(-4) = \frac{-4}{-4+2} = 2 \). If \( x = -1 \): \( f(-1) = \frac{-1}{-1+2} = -1 \). If \( x = 1 \): \( f(1) = \frac{1}{1+2} \approx 0.33 \). These points can be plotted along with the intercepts to get a more complete view of the function's graph. Don鈥檛 forget to plot the points carefully and draw the curve approaching the asymptotes.
