Problem 45 - 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. $$f(x)=\frac{3 x}{(x+1)(x-2)}$$

Step-by-Step Solution

Answer

The vertical asymptotes are at \(x = -1\) and \(x = 2\); the horizontal asymptote is \(y = 0\). The graph has x- and y-intercept at \((0,0)\).

Step 1: Identify Vertical Asymptotes

Vertical asymptotes occur where the denominator of a rational function is zero and the numerator is not zero. Factor the denominator: \[(x+1)(x-2) = 0\]Solve for the vertical asymptotes by setting each factor to zero:- \(x+1=0\) gives \(x = -1\)- \(x-2=0\) gives \(x = 2\)Thus, there are vertical asymptotes at \(x = -1\) and \(x = 2\).

Step 2: Identify Horizontal Asymptote

To find the horizontal asymptote, consider the degrees of the polynomial in the numerator and the polynomial in the denominator:- The degree of the numerator \(3x\) is 1.- The degree of the denominator \((x+1)(x-2) = x^2 - x - 2\) is 2.When the degree of the numerator is less than the degree of the denominator, the horizontal asymptote is \(y = 0\).

Step 3: Find the x-intercept

The x-intercept occurs when the numerator of the rational function is zero. Set the numerator equal to zero:\[3x = 0\]Solve for \(x\):\[x = 0\]Thus, there is one x-intercept at \(x = 0\) (the point \((0,0)\)).

Step 4: Find the y-intercept

The y-intercept occurs when \(x = 0\). Substitute \(x = 0\) into the function:\[f(0) = \frac{3(0)}{(0+1)(0-2)} = \frac{0}{0-2} = 0\]Thus, the y-intercept is \(f(0) = 0\), confirming the point \((0, 0)\).

Step 5: Analyze the Behavior Near Asymptotes

Check the sign of the function values around the vertical asymptotes:- Test \(x = -2\), close to \(x = -1\): \(\frac{3(-2)}{((-2)+1)((-2)-2)} = \frac{-6}{-3} = 2\) (positive)- Test \(x = 1\), close to \(x = 2\): \(\frac{3(1)}{(1+1)(1-2)} = \frac{3}{-2} = -1.5\) (negative) The sign changes around the vertical asymptotes, confirming the vertical asymptotes' orientation.

Step 6: Sketch the Graph

Now, sketch the rational function based on the information:1. Draw vertical asymptotes at \(x = -1\) and \(x = 2\).2. Draw the horizontal asymptote as the line \(y=0\).3. Include the intercepts at \((0, 0)\).4. Use the sign changes around the asymptotes to determine the curve's direction.The curve approaches but doesn’t cross the asymptotes, moves away from \(x = -1\) reaching the y-axis, and reverts towards \(x = 2\). It stays below the x-axis after \(x=0\) until it reaches the vertical asymptote at \(x=2\). Lastly, make sure the function behaves according to the asymptotes as \(x\) approaches infinity.

Key Concepts

Vertical Asymptotes

Vertical asymptotes are critical features when graphing a rational function. They appear as vertical lines on a graph where the function heads towards infinity. In simpler terms, these are points that the function approaches but never really touches or crosses.

To find vertical asymptotes, you need to check where the denominator equals zero, while the numerator remains non-zero. For the function \( f(x) = \frac{3x}{(x+1)(x-2)} \), factor the denominator to identify these points. Set the factors \((x + 1)\) and \((x - 2)\) each to zero:

• Solving \(x + 1 = 0\) gives \(x = -1\).
• Solving \(x - 2 = 0\) gives \(x = 2\).

These calculations reveal two vertical asymptotes located at \(x = -1\) and \(x = 2\). Remember, the graph will approach these lines but will never intersect them.

Horizontal Asymptotes

Horizontal asymptotes describe the behavior of a graph as \(x\) approaches positive or negative infinity. Essentially, they indicate where the graph levels off.

To determine horizontal asymptotes in a rational function, compare the degrees of the numerator and the denominator:

• The degree of the numerator is the highest power of \(x\) in the numerator. In \(3x\), the degree is 1.
• The degree of the denominator is found in \((x+1)(x-2)\), which expands to \(x^2 - x - 2\). The degree is 2.

Here, the degree of the numerator (1) is less than that of the denominator (2). When this is the case, the horizontal asymptote is \(y = 0\), meaning the graph approaches the x-axis as \(x\) moves towards infinity.

The horizontal asymptote provides a simple visual checkpoint, helping you understand how the function behaves at extreme ends.

Intercept Analysis

Intercepts are vital for understanding where the graph of a function crosses the x or y axes. They offer another critical insight into the structure of a rational function.

**X-intercept**
To find the x-intercept, set the numerator equal to zero and solve for \(x\). For \(f(x) = \frac{3x}{(x+1)(x-2)} \):

• Solve \(3x = 0\) to find \(x = 0\).

The x-intercept is at point \((0, 0)\).

**Y-intercept**
The y-intercept can be determined by setting \(x=0\) in the entire function:

• \(f(0) = \frac{3(0)}{(0+1)(0-2)} = 0\).

The y-intercept is also at \((0, 0)\).

Keep in mind, a rational function might have limited intercepts due to the nature of its components, but they are still crucial for plotting accurate graphs.