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Magazine Chapter 1: Problem 17
In Exercises 5鈥�30, determine an appropriate viewing window for the given function and use it to display its graph. $$ y=\frac{x+3}{x+2} $$
Short Answer
Expert verified
Viewing window: x from -6 to 2, y from -2 to 3.
Step by step solution
01
Identify Domain Restrictions
The function \( y = \frac{x+3}{x+2} \) is undefined where the denominator is zero. Set the denominator \( x+2 = 0 \) and solve for \( x \). The solution \( x = -2 \) indicates a vertical asymptote at \( x = -2 \). The domain of the function is all real numbers except \( x = -2 \).
02
Determine Asymptotes
Aside from the vertical asymptote at \( x = -2 \), determine the horizontal asymptote by dividing the coefficients of the leading terms in the numerator and the denominator, since they have the same degree. The horizontal asymptote is \( y = 1 \).
03
Determine Behavior at Intercepts
Find the x-intercept by setting the numerator equal to zero: \( x+3 = 0 \). This gives \( x = -3 \). The y-intercept occurs when \( x = 0 \), which gives \( y = \frac{3}{2} = 1.5 \). Use these points as reference points for setting up the viewing window.
04
Choose a Viewing Window
Based on the location and behavior of the asymptotes and intercepts, a reasonable viewing window for the function is from \( x = -6 \) to \( x = 2 \) for the x-axis to include the intercepts and the asymptote behavior; and from \( y = -2 \) to \( y = 3 \) on the y-axis to see the horizontal asymptote and intercepts clearly.
05
Confirm with Graph
Graph the function using the selected window: \( x = -6 \, ext{to} \, 2 \), \( y = -2 \, ext{to} \, 3 \). Ensure the features (asymptotes and intercepts) are visible and match the analyses performed. Adjust the window slightly if necessary to better visualize critical points or behavior.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Domain Restrictions
Domain restrictions in a function occur when the denominator equals zero because division by zero is undefined in mathematics. In the function \( y = \frac{x+3}{x+2} \), any value of \( x \) that makes the denominator zero must be excluded from the domain. To find this, set \( x+2 = 0 \) and solve for \( x \), which gives \( x = -2 \). Consequently, the domain of the function is all real numbers except \( x = -2 \).
This means the graph will have a vertical asymptote at \( x = -2 \), and the function will approach but never meet this line. Understanding these restrictions is vital for correctly plotting the function.
This means the graph will have a vertical asymptote at \( x = -2 \), and the function will approach but never meet this line. Understanding these restrictions is vital for correctly plotting the function.
Vertical Asymptote
A vertical asymptote represents a value of \( x \) where the function becomes undefined, and the graph heads towards infinity. For the function \( y = \frac{x+3}{x+2} \), the vertical asymptote occurs at \( x = -2 \), as previously identified by the domain restriction.
- A vertical asymptote does not cross the graph.
- The function will approach this line where the graph appears to go "up" or "down" indefinitely.
- It represents a significant point to consider when choosing a viewing window.
Horizontal Asymptote
A horizontal asymptote is a horizontal line that the graph of a function approaches as \( x \) goes to infinity or negative infinity. To determine it for \( y = \frac{x+3}{x+2} \), you look at the coefficients of the highest degree terms in the numerator and denominator, which are both 1. Therefore, the horizontal asymptote is \( y = 1 \).
- This shows the behavior of the function as \( x \) becomes very large or very small.
- Unlike vertical asymptotes, a graph can cross its horizontal asymptote.
- Horizontal asymptotes provide useful information about the end behavior of the graph, crucial for defining a good viewing window on the y-axis.
Graph Intercepts
Intercepts are specific points where the graph crosses the x- and y-axes. They serve as essential guides for understanding and plotting the graph thoroughly. For the equation \( y = \frac{x+3}{x+2} \):
- The x-intercept occurs where \( y = 0 \). Solving for \( x \) by setting the numerator zero, \( x+3 = 0 \) gives \( x = -3 \).
- The y-intercept is where \( x = 0 \). Substitute \( x \) to get \( y = \frac{3}{2} = 1.5 \).
