Problem 57 - A Graphical Approach to Precalculus with Limits

Question

Sketch a graph of rational function. Your graph should include all asymptotes. Do not use a calculator. $$f(x)=\frac{9 x^{2}-1}{x^{2}-4}$$

Step-by-Step Solution

Answer

Vertical asymptotes at \( x = \pm 2 \), horizontal asymptote at \( y = 9 \). Intercepts at \((0, \frac{1}{4}), (\frac{1}{3}, 0), (-\frac{1}{3}, 0)\).

Step 1: Identify Vertical Asymptotes

Vertical asymptotes occur where the denominator equals zero, and the numerator does not equal zero. For the function \( f(x) = \frac{9x^2 - 1}{x^2 - 4} \), set the denominator equal to zero: \[ x^2 - 4 = 0 \] Solving for \( x \), we find \( x^2 = 4 \) or \( x = \pm 2 \). Therefore, the vertical asymptotes are at \( x = 2 \) and \( x = -2 \).

Step 2: Identify Horizontal Asymptotes

Horizontal asymptotes depend on the degrees of the polynomial in the numerator and denominator. Both the numerator \( 9x^2 - 1 \) and denominator \( x^2 - 4 \) are degree 2 polynomials, and their leading coefficients are 9 and 1, respectively. Thus, the horizontal asymptote is \( y = \frac{9}{1} = 9 \).

Step 3: Find Intercepts

Finding intercepts helps to understand the behavior of the graph. To find the y-intercept, set \( x = 0 \) in \( f(x) \): \[ f(0) = \frac{9(0)^2 - 1}{0^2 - 4} = \frac{-1}{-4} = \frac{1}{4} \]. So, the y-intercept is \((0, \frac{1}{4})\). To find x-intercepts, set the numerator equal to zero and solve for \( x \): \[ 9x^2 - 1 = 0 \]\[ 9x^2 = 1 \]\[ x^2 = \frac{1}{9} \]\[ x = \pm \frac{1}{3} \]. So, the x-intercepts are at \( \left( \frac{1}{3}, 0 \right) \) and \( \left( -\frac{1}{3}, 0 \right) \).

Step 4: Sketch the Graph Using Asymptotes and Intercepts

Begin by drawing the vertical asymptotes as dotted lines at \( x = 2 \) and \( x = -2 \), and the horizontal asymptote as a dotted line at \( y = 9 \). Plot the intercepts: the y-intercept at \( (0, \frac{1}{4}) \) and x-intercepts at \( \left( \frac{1}{3}, 0 \right) \) and \( \left( -\frac{1}{3}, 0 \right) \). The graph approaches the vertical asymptotes, going towards infinity or negative infinity as \( x \) nears \( 2 \) or \( -2 \), and approaches \( y = 9 \) as \( x \) approaches infinity or negative infinity.

Key Concepts

Vertical Asymptotes

Vertical asymptotes are a critical feature in the graph of a rational function. They indicate the x-values where the function's value becomes unbounded, meaning neither positive nor negative numbers can represent it. This occurs when the denominator of the rational function equals zero, making the function undefined. For example, considering the given function: \[ f(x) = \frac{9x^2 - 1}{x^2 - 4} \]The denominator is \( x^2 - 4 \). Setting this equal to zero gives us:\[ x^2 - 4 = 0 \]When solved, it provides the values \( x = \pm 2 \). Therefore, these values correspond to vertical asymptotes at \( x = 2 \) and \( x = -2 \). Points to Remember:

• Vertical asymptotes occur where the denominator is zero and the numerator isn't zero.
• The graph will approach these lines but will never touch them.

They symbolize a boundary in the function where the value tends to positive or negative infinity.

Horizontal Asymptotes

Horizontal asymptotes signify the behavior of a rational function as the x-values grow very large (positively or negatively). Unlike vertical asymptotes, they indicate the y-value the function approaches but does not necessarily cross or remain constant at when \( x \) is extremely high or low.For a function \( f(x) = \frac{9x^2 - 1}{x^2 - 4} \), the degree of the numerator and denominator is crucial in identifying the horizontal asymptote. Both the numerator (\( 9x^2 - 1 \)) and the denominator (\( x^2 - 4 \)) are degree two polynomials. The leading coefficient rule applies here. As both polynomials are of the same degree, divide the leading coefficients: \[ \frac{9}{1} = 9 \]This provides a horizontal asymptote at \( y = 9 \).Insights:

• If the degree of the numerator is less than the denominator, \( y = 0 \) is the horizontal asymptote.
• If the degrees are the same, divide the leading coefficients for the asymptote.

The graph's behavior approaches this horizontal line at extreme values of \( x \), mirroring steady state or saturation in real-world scenarios.

Intercepts

Intercepts help us mark points where the graph of a function crosses the axes, offering valuable clues about its behavior.**Y-Intercept**: This is where the graph crosses the y-axis. To find the y-intercept of the rational function, substitute \( x = 0 \) in \( f(x) \):\[ f(0) = \frac{9(0)^2 - 1}{0^2 - 4} = \frac{-1}{-4} = \frac{1}{4} \].Thus, the y-intercept is at \( (0, \frac{1}{4}) \).**X-Intercepts**: These are the points where the graph crosses the x-axis; found by setting the numerator to zero:\[ 9x^2 - 1 = 0 \]Solving for \( x \), yields:\[ 9x^2 = 1 \]\[ x^2 = \frac{1}{9} \]\[ x = \pm \frac{1}{3} \].Hence, the x-intercepts are \( \left( \frac{1}{3}, 0 \right) \) and \( \left( -\frac{1}{3}, 0 \right) \).Considerations:

• Y-intercepts occur where \( x = 0 \).
• X-intercepts are solutions to the numerator equated to zero.

Intercepts give starting points for graph sketching and illuminate changes in the function's sign.