Problem 17 - Thomas Calculus

Question

In Exercises 5–30, determine an appropriate viewing window for the given function and use it to display its graph. $$ y=\frac{x+3}{x+2} $$

Step-by-Step Solution

Answer

Viewing window: x from -6 to 2, y from -2 to 3.

Step 1: 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 \( x = -2 \).

Step 2: 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 \( y = 1 \).

Step 3: 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.

Step 4: 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.

Step 5: 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.

Key Concepts

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.

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.

Thus, knowing the location of the vertical asymptote helps in setting the boundaries to view the essential features of the graph.

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.

Recognizing where the horizontal asymptote lies helps to ensure it's visible when selecting the graph's viewing dimensions.

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 \( y = \frac{3}{2} = 1.5 \).

These intercepts are crucial markers on the graph, helping to calibrate the viewing window. Knowing their exact location ensures these important crossings are visible, aiding in better visualization of the function's complete behavior.