Problem 15 - Algebra and Trigonometry with Analytic Geometry

Question

Sketch the graph of \(f\). $$ f(x)=\frac{x-2}{x^{2}-x-6} $$

Step-by-Step Solution

Answer

The function \(f(x)\) has vertical asymptotes at \(x = -2\) and \(x = 3\), a horizontal asymptote at \(y = 0\), with an x-intercept at \(x = 2\) and a y-intercept at \(y = \frac{1}{3}\).

Step 1: Factor the Denominator

First, factor the quadratic expression in the denominator, which is \(x^2 - x - 6\). We can factor it as follows:\[x^2 - x - 6 = (x - 3)(x + 2)\]Thus, the function can be rewritten as:\[f(x) = \frac{x-2}{(x-3)(x+2)}\]

Step 2: Identify the Vertical Asymptotes

Vertical asymptotes occur where the denominator is zero (and the numerator isn't zero at the same point). Set each factor in the denominator equal to zero:\[x - 3 = 0 \quad \text{or} \quad x + 2 = 0\]Thus, \(x = 3\) and \(x = -2\) are the vertical asymptotes.

Step 3: Determine the Horizontal Asymptote

The horizontal asymptote is found by comparing the degrees of the numerator and the denominator. In this function, \(f(x) = \frac{x-2}{x^2-x-6}\), the degree of the numerator is 1, and the degree of the denominator is 2. Since the numerator degree is less than the denominator degree, the horizontal asymptote is:\[y = 0\]

Step 4: Determine Critical Points and Intercepts

To find the x-intercept, set the numerator equal to zero:\[x - 2 = 0 \Rightarrow x = 2\]The y-intercept occurs when \(x = 0\):\[f(0) = \frac{0 - 2}{0^2 - 0 - 6} = \frac{-2}{-6} = \frac{1}{3}\]

Step 5: Analyze the Function Behavior

Analyze the sign changes around the asymptotes and intercepts. Check how \(f(x)\) behaves as \(x\to 3\) from both sides, as \(x\to -2\), and as \(x\to \pm \infty\). You can choose test points in each interval defined by \(x = -2,\ x = 2,\ x = 3\).

Step 6: Sketch the Graph

Using the identified asymptotes, intercepts, and behavior analysis, sketch the graph:- The graph approaches vertical asymptotes at \(x = -2\) and \(x = 3\).- The graph will have a horizontal asymptote at \(y = 0\).- Mark the intercepts at \((2, 0)\) and \((0, \frac{1}{3})\).- Analyze function values between and beyond these critical points to sketch continuous curves approaching the asymptotes.

Key Concepts

Graphing Rational Functions

Graphing rational functions may feel a bit tricky at first, but it's all about understanding the behavior and characteristics of the function. A rational function is the division of two polynomial expressions. In our case, we have:

• The numerator: \(x - 2\)
• The denominator: \((x - 3)(x + 2)\)

Start by identifying key features that influence the graph's shape. This includes vertical and horizontal asymptotes, intercepts, and critical points. By systematically analyzing these elements, you can sketch the function's behavior and plot its trajectory.

Vertical Asymptotes

Vertical asymptotes occur at values of \(x\) where the function's denominator equals zero and the numerator does not. These are points where the graph heads towards infinity, meaning the function doesn't approach any specific finite value.
For the given function, we set the denominator equal to zero:

• \(x - 3 = 0\) gives \(x = 3\)
• \(x + 2 = 0\) gives \(x = -2\)

As a result, there are vertical asymptotes at \(x = 3\) and \(x = -2\). These tell you where the function shoots off towards positive or negative infinity, creating barriers that the graph can never cross. Be sure to note these locations when sketching the graph, as they define regions in which the function's values rapidly increase or decrease.

Horizontal Asymptotes

Horizontal asymptotes indicate the end behavior of a function as \(x\) approaches positive or negative infinity. It's all about comparing the degrees of the numerator and the denominator.
In this function, the numerator has a degree of 1 (\(x\)), while the denominator has a degree of 2 (the highest degree term, \(x^2\)). When the degree of the denominator is greater than that of the numerator, the horizontal asymptote is:

• \(y = 0\)

This tells us how the function will behave as \(x\) moves towards infinity. The graph will approach the \(x\)-axis but will never actually touch or cross it at the far ends. Understanding this gives you clues about the horizontal alignment of the graph.

Intercepts

Intercepts help in determining where the graph of the function crosses the axes. They provide anchor points within the sketch of the graph.

• **X-intercept:** This is found by setting the numerator equal to zero and solving for \(x\).
  From \(x - 2 = 0\), we find the x-intercept at \((2, 0)\).


• **Y-intercept:** This is determined by evaluating the function at \(x = 0\).
  \(f(0) = \frac{0 - 2}{0^2 - 0 - 6} = \frac{1}{3}\).
  Thus, the y-intercept is \((0, \frac{1}{3})\).

These intercepts give you a solid starting point for drawing the graph. They show where the curve crosses the axes, aiding in the visualization of the function's path across the coordinate plane.