Problem 18 - Precalculus with Calculus Previews

Question

Find the vertical and horizontal asymptotes for the graph of the given rational function. Find \(x\) - and \(y\) -intercepts of the graph. Sketch the graph of \(f\). $$ f(x)=\frac{1}{x^{2}-2 x-8} $$

Step-by-Step Solution

Answer

Vertical asymptotes at \(x = 4\) and \(x = -2\), horizontal asymptote at \(y = 0\), \(y\)-intercept at \((0, -\frac{1}{8})\), no \(x\)-intercepts.

Step 1: Identify Vertical Asymptotes

To find the vertical asymptotes, set the denominator equal to zero and solve for \(x\): \[ x^2 - 2x - 8 = 0 \] Factor the quadratic: \[ (x - 4)(x + 2) = 0 \] Setting each factor to zero gives the vertical asymptotes at \(x = 4\) and \(x = -2\).

Step 2: Identify Horizontal Asymptote

For the rational function \(f(x)=\frac{1}{x^2-2x-8}\), since the degree of the polynomial in the numerator is less than the degree of the polynomial in the denominator, there is a horizontal asymptote at \(y = 0\).

Step 3: Find x-intercepts

To find the \(x\)-intercepts, set the numerator equal to zero because the function is \(f(x) = \frac{1}{x^2 - 2x - 8}\). Since the numerator is constant \(1\), it never reaches zero, so there are no \(x\)-intercepts.

Step 4: Find y-intercepts

To find the \(y\)-intercept, evaluate \(f(0)\): \[ f(0) = \frac{1}{0^2 - 2 \times 0 - 8} = \frac{1}{-8} = -\frac{1}{8} \] Thus, the \(y\)-intercept is \((0, -\frac{1}{8})\).

Step 5: Sketch the Graph

Using the vertical asymptotes \(x = 4\) and \(x = -2\), the horizontal asymptote \(y = 0\), and the \(y\)-intercept \((0, -\frac{1}{8})\), sketch the graph. The graph will approach the vertical asymptotes but never touch or cross them. Similarly, it will approach the horizontal asymptote as \(x\) goes to \(\pm \infty\).

Key Concepts

Vertical Asymptotes

Vertical asymptotes occur where the denominator of a rational function is zero, causing the function to become undefined. For the function \( f(x) = \frac{1}{x^2 - 2x - 8} \), we find vertical asymptotes by setting the denominator equal to zero and solving:

• First, we solve the equation \( x^2 - 2x - 8 = 0 \).
• This quadratic equation factors into \((x - 4)(x + 2) = 0\).
• By setting each factor equal to zero, we get \( x = 4 \) and \( x = -2 \).

The function doesn’t exist at these \( x \)-values, marking the locations of the vertical asymptotes. Hence, the graph approaches these lines, but will never intersect them. This behavior is crucial for understanding how the function behaves near these points.

Horizontal Asymptotes

Horizontal asymptotes of a rational function describe its end behavior, indicating the value that the function approaches as \( x \) tends towards infinity or negative infinity. To find the horizontal asymptote of \( f(x) = \frac{1}{x^2 - 2x - 8} \), we compare the degrees of the numerator and the denominator:

• The numerator is \(1\), a constant, meaning its degree is \(0\).
• The denominator is \(x^2 - 2x - 8\), with a degree of \(2\).

Since the degree of the numerator is less than the degree of the denominator, the horizontal asymptote is \( y = 0 \). This tells us that as \( x \) becomes very large or very small, the value of \( f(x) \) gets closer and closer to 0, never actually reaching it.

Intercepts of Rational Functions

Finding the intercepts of a rational function helps in understanding where the graph crosses the axes.**X-intercepts:**- These occur where the graph crosses the x-axis, or \( f(x) = 0 \). - For \( f(x) = \frac{1}{x^2 - 2x - 8} \), the numerator is \(1\). Since \(1\) is never zero, there are no x-intercepts.**Y-intercepts:**- These are found where the graph crosses the y-axis, or where \( x = 0 \).- Plugging \( x = 0 \) into the function gives \( f(0) = \frac{1}{0^2 - 2 \cdot 0 - 8} = -\frac{1}{8} \).- Therefore, the y-intercept is at \((0, -\frac{1}{8})\). These intercepts are important as they provide key points that help in sketching the graph accurately.