Problem 2
Question
Match the rational function in Column I with the appropriate description in Column II. Choices in Column II can be used only once. Do not use a calculator. \(\mathbf{I}\) \(f(x)=\frac{x+10}{x+2}\) \(\mathbf{II}\) A. The \(x\) -intercept is \((-3,0)\) B. The \(y\) -intercept is \((0,5)\) C. The horizontal asymptote is \(y=4\) D. The vertical asymptote is \(x=-1\) E. There is a hole in its graph at \(x=-4\) F. The graph has an oblique asymptote. G. The \(x\) -axis is its horizontal asymptote. H. The \(y\) -axis is its vertical asymptote.
Step-by-Step Solution
Verified Answer
B: The y-intercept is (0,5).
1Step 1: Determine x-intercepts
To find the \(x\)-intercept of \(f(x)=\frac{x+10}{x+2}\), set the numerator equal to zero: \(x+10=0\). This gives \(x=-10\). Therefore, the \(x\)-intercept is \((-10, 0)\). Description A does not match.
2Step 2: Determine y-intercepts
To find the \(y\)-intercept, substitute \(x = 0\) into the function: \(f(0) = \frac{0+10}{0+2} = \frac{10}{2} = 5\). Thus, the \(y\)-intercept is \((0, 5)\). This matches description B.
3Step 3: Determine vertical asymptotes
Vertical asymptotes occur where the denominator is zero and the numerator is not zero. Set the denominator \(x+2=0\), giving \(x=-2\), so the vertical asymptote is \(x=-2\). None of the descriptions in Column II match this result.
4Step 4: Determine horizontal asymptotes
For the rational function \(\frac{x+10}{x+2}\), the degrees of the numerator and the denominator are equal. The horizontal asymptote is found by dividing the leading coefficients: \(y=\frac{1}{1}=1\). None of the descriptions in Column II directly match this result.
5Step 5: Conclusion
Based on the analysis, only the \(y\)-intercept description (B: \((0,5)\)) correctly matches the function \(f(x)=\frac{x+10}{x+2}\). No exact matches for vertical or horizontal asymptotes are present in the given descriptions, and the \(x\)-intercept was not listed either.
Key Concepts
AsymptotesInterceptsGraph Analysis
Asymptotes
Asymptotes are essential features of rational functions. They represent lines that the graph of the function approaches but never actually touches. There are two main types: vertical and horizontal.
- Vertical Asymptotes: These occur when the denominator of the rational function equals zero, leading to undefined values at certain points. For the function \( f(x) = \frac{x+10}{x+2} \), we set the denominator \( x+2=0 \), which results in \( x=-2 \). This means there is a vertical asymptote at \( x=-2 \), indicating that as the graph nears \( x=-2 \), it heads toward infinity or negative infinity.
- Horizontal Asymptotes: When both the numerator and denominator of a rational function have the same degree, the horizontal asymptote can be found by dividing the leading coefficients of these terms. In this case, as both the numerator and denominator have a degree of 1, the horizontal asymptote is the line \( y=1 \).
Intercepts
Intercepts are points on the graph where the rational function crosses the axes. These are crucial for understanding the starting and stopping points of a graph in a visual manner.
- \(x\)-Intercepts: To find the \( x \)-intercept, set the numerator equal to zero. For our function \( f(x) = \frac{x+10}{x+2} \), solving \( x+10=0 \) gives \( x=-10 \). Thus the \( x \)-intercept is \((-10, 0)\). It represents a point where the graph crosses the \( x \)-axis.
- \(y\)-Intercepts: These are found by setting \( x=0 \) in the function. Substituting into the function gives \( f(0) = \frac{10}{2} = 5 \). Therefore, the \( y \)-intercept is \((0, 5)\). This represents a point where the graph crosses the \( y \)-axis.
Graph Analysis
Graph analysis of rational functions involves understanding both intercepts and asymptotes, as they are critical in plotting and visualizing the graph of the function.
- Understanding the Shape: Combining both intercepts and asymptotes provides a skeleton for the graph. The intercepts tell us where the graph will cross the axes, while the asymptotes indicate where the graph trends toward infinity or stabilizes.
- Behavior Near Asymptotes: As the function approaches the vertical asymptote, the graph will ascend or descend steeply, depending on the function's behavior. Near horizontal asymptotes, the graph will flatten out, becoming nearly horizontal as \( x \) extends toward positive or negative infinity.
- Overall Appearance: Using your known intercepts and asymptotes, sketch the general shape of the graph. Notably, the function \( f(x) = \frac{x+10}{x+2} \) has a vertical asymptote at \( x = -2 \) and a horizontal asymptote at \( y = 1 \). This yields a graph that curves around \( x = -2 \) and eventually aligns to a stable horizontal line at \( y = 1 \).
Other exercises in this chapter
Problem 2
Evaluate each expression. $$-\sqrt[3]{64}$$
View solution Problem 2
In Exercises begin by drawing a rough sketch to determine the number of real solutions for the equation \(y_{1}=y_{2}\) Then solve this equation by hand. Give t
View solution Problem 2
Provide a short answer to each question. Do not use a calculator. What is the domain of \(f(x)=\frac{1}{x^{2}}\) ? What is its range?
View solution Problem 3
Evaluate each expression. $$\sqrt[5]{-32}$$
View solution