Problem 47 - A Graphical Approach to College Algebra

Question

Sketch a graph of each rational function. Your graph should include all asymptotes. Do not use a calculator. $$f(x)=\frac{x}{x^{2}-9}$$

Step-by-Step Solution

Answer

Graph has vertical asymptotes at \( x = -3 \) and \( x = 3 \), and a horizontal asymptote at \( y = 0 \), with the origin as an intercept.

Step 1: Identify Domain Restrictions

The function given is \( f(x) = \frac{x}{x^2 - 9} \). Start by identifying the domain of \( f(x) \). The denominator cannot be zero, so set \( x^2 - 9 = 0 \) and solve for \( x \). This results in \( x = \pm 3 \). Therefore, the domain of the function is all real numbers except \( x = 3 \) and \( x = -3 \).

Step 2: Determine Vertical Asymptotes

Vertical asymptotes occur where the denominator is zero and the numerator is not zero; thus, look at the values found in Step 1. The vertical asymptotes are at \( x = -3 \) and \( x = 3 \) since the numerator \( x \) is not zero at these points.

Step 3: Find Horizontal Asymptotes

Horizontal asymptotes depend on the degrees of the numerator and denominator. The degree of the numerator \( x \) is 1, and the degree of the denominator \( x^2 - 9 \) is 2. The degree of the numerator is less than that of the denominator, indicating a horizontal asymptote at \( y = 0 \).

Step 4: Analyze Critical Points and Intercepts

Find the \( x \)-intercept by setting \( f(x) = 0 \), which occurs when the numerator is zero. Since the numerator is \( x \), the \( x \)-intercept is at \( x = 0 \). Also, check the \( y \)-intercept which is the same point (\( 0, 0 \)), as \( f(0) = 0 \).

Step 5: Sketch the Graph

Using the information gathered: plot vertical asymptotes as dashed lines at \( x = -3 \) and \( x = 3 \). Draw a horizontal asymptote along \( y = 0 \). Mark the intercept \( (0,0) \). Sketch the graph in regions around the asymptotes, respecting the intercept and asymptotes. The graph approaches the horizontal asymptote as \( x \) moves towards \( \pm \infty \) and breaks at \( x = -3 \) and \( x = 3 \).

Key Concepts

Domain Restrictions

When dealing with rational functions, like \( f(x) = \frac{x}{x^2 - 9} \), identifying domain restrictions is important. A rational function's domain consists of all real numbers that do not make the denominator zero. To find these restrictions, solve the equation \( x^2 - 9 = 0 \). This leads to \(x = \pm 3\). As a result, the domain for this function is all real numbers except \( x = 3 \) and \( x = -3 \). Always remember:

• If the denominator equals zero, the function is undefined at that point.
• Exclude such values from the domain.

Vertical Asymptotes

Vertical asymptotes occur in rational functions where the denominator equals zero, but the numerator is not zero. This occurs because the function approaches infinity near these points. For \( f(x) = \frac{x}{x^2 - 9} \), the vertical asymptotes are found at the values \( x = 3 \) and \( x = -3 \). The reason is simple: the denominator becomes zero and since the numerator \( x \) does not equal zero at these points, both create vertical asymptotes. Here's a little tip:

• Vertical asymptotes divide the graph into separate regions.
• They represent divisions beyond which the function’s value dramatically changes.

Horizontal Asymptotes

Horizontal asymptotes tell us the behavior of the function as \( x \) goes to infinity or negative infinity. To find horizontal asymptotes, compare the degrees of the numerator and denominator.In \( f(x) = \frac{x}{x^2 - 9} \), the degree of the numerator is 1, and the denominator is 2. Because the numerator's degree is less, this indicates there is a horizontal asymptote at \( y = 0 \). This implies:

• The function values become closer to zero as \( x \) becomes very large or very small.
• Provides a boundary the curve approaches but never touches as it extends horizontally.

Intercepts

Intercepts are the points where the graph crosses the axes, important for understanding the function's behavior.The \( x \)-intercept of a function occurs where the function equals zero, meaning the numerator should be zero. For \( f(x) = \frac{x}{x^2 - 9} \), the \( x \)-intercept is at \( x = 0 \) since \( x = 0 \) makes the numerator zero.Similarly, the \( y \)-intercept occurs where \( x = 0 \). Substituting \( x = 0 \) in \( f(x) \) results in \( f(0) = 0 \), which means the \( y \)-intercept is also at \( (0,0) \).Key insights:

• Intercepts provide starting points for sketching the graph.
• They help in shaping the curve and understanding its motion across axes.