Problem 35 - 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. $$f(x)=\frac{3 x}{(x+1)(x-2)}$$

Step-by-Step Solution

Answer

Asymptotes: \(x=-1, x=2\) (vertical), \(y=0\) (horizontal). X-/Y-intercept: \((0,0)\).

Step 1: Identify the Vertical Asymptotes

Find the values of \(x\) that make the denominator zero. The denominator is \((x+1)(x-2)\). Setting each factor equal to zero gives us:- \(x + 1 = 0\) leading to \(x = -1\)- \(x - 2 = 0\) leading to \(x = 2\)Thus, the vertical asymptotes are at \(x = -1\) and \(x = 2\).

Step 2: Identify the Horizontal Asymptote

Compare the degrees of the numerator and the denominator. The degree of the numerator \(3x\) is 1, and the degree of the denominator \((x+1)(x-2)\) is 2.Since the degree of the numerator is less than the degree of the denominator, the horizontal asymptote is \(y = 0\).

Step 3: Finding the X-Intercept

To find the x-intercept, set the numerator equal to zero: \(3x = 0\).Solving for \(x\), we get \(x = 0\).So, the x-intercept of the function is at the point \((0, 0)\).

Step 4: Finding the Y-Intercept

To find the y-intercept, substitute \(x = 0\) into the function: \[f(0) = \frac{3 \times 0}{(0+1)(0-2)} = \frac{0}{-2} = 0\]So, the y-intercept is also at the point \((0, 0)\).

Step 5: Analyze the Sign Changes

Choose test points in each interval divided by the vertical asymptotes and the x-intercept to determine the sign of \(f(x)\):- Interval \((-\infty, -1)\), test \(x = -2\): \(f(-2) = \frac{3(-2)}{(-2+1)(-2-2)} = \frac{-6}{-12} > 0\), \(f(x) > 0\).- Interval \((-1, 0)\), test \(x = -0.5\): \(f(-0.5)= \frac{3(-0.5)}{(-0.5+1)(-0.5-2)} = \frac{-1.5}{1.25} < 0\), \(f(x) < 0\).- Interval \((0, 2)\), test \(x = 1\): \(f(1) = \frac{3(1)}{(1+1)(1-2)} = \frac{3}{-2} < 0\), \(f(x) < 0\).- Interval \((2, \infty)\), test \(x = 3\): \(f(3) = \frac{3(3)}{(3+1)(3-2)} = \frac{9}{4} > 0\), \(f(x) > 0\).

Step 6: Sketch the Graph

Using the information gathered:- The vertical asymptotes are at \(x = -1\) and \(x = 2\).- The horizontal asymptote is \(y = 0\).- The function crosses the x-axis at \(x = 0\).- Consider the sign changes over the intervals: - \(f(x) > 0\) for \(x < -1\) and \(x > 2\). - \(f(x) < 0\) for \(-1 < x < 0\) and \(0 < x < 2\).Sketch the curve by approaching the asymptotes as guides while transitioning through the intercepts and sign changes. The function will not touch the horizontal asymptote at \(y = 0\), except as \(x\) approaches infinity.

Key Concepts

Vertical Asymptotes

Vertical asymptotes are important features of rational functions, and they occur when the denominator equals zero—making the function undefined at those points. For the function \(f(x) = \frac{3x}{(x+1)(x-2)}\), the vertical asymptotes can be found by setting each factor in the denominator to zero.

• For \(x + 1 = 0\), solve to find \(x = -1\).
• For \(x - 2 = 0\), solve to find \(x = 2\).

This means there are vertical asymptotes at \(x = -1\) and \(x = 2\). When graphing, the function's curve will approach these vertical lines but will never quite touch them. Understanding vertical asymptotes is crucial for sketching and analyzing the behavior of rational functions.

Horizontal Asymptotes

Horizontal asymptotes help indicate the end behaviour of a rational function as \(x\) approaches infinity or negative infinity. To determine the horizontal asymptote for \(f(x) = \frac{3x}{(x+1)(x-2)}\), compare the degrees of the numerator and denominator.

• The numerator \(3x\) is of degree 1.
• The denominator \((x+1)(x-2)\) simplifies to a quadratic form, having degree 2.

Since the degree of the numerator is less than the degree of the denominator, the horizontal asymptote is \(y = 0\). This signifies that as \(x\) grows large or very negative, the value of \(f(x)\) will get closer and closer to zero, without ever actually reaching it. Horizontal asymptotes provide a way to predict what happens to the function at extreme values of \(x\).

Graphing Rational Functions

Graphing rational functions involves several steps to capture their key characteristics, including asymptotes and intercepts. 1. **Identify Asymptotes:** As determined, \(x = -1\) and \(x = 2\) are the vertical asymptotes, and \(y = 0\) is the horizontal asymptote.2. **Find Intercepts:** The \(x\)-intercept can be found by setting the numerator equal to zero. Here, \(3x = 0\), leading to \(x = 0\), placing the intercept at (0,0). The \(y\)-intercept is found by evaluating \(f(0)\), which is also (0,0).3. **Divide Number Line and Test Signs:** Use vertical asymptotes and intercepts to divide the graph into intervals and test for positivity or negativity.When these elements combine, they shape how you sketch the curve to respect the behavior dictated by the asymptotes and intercepts.

X-Intercepts

Finding the \(x\)-intercepts of a rational function involves setting the numerator equal to zero to identify where the graph crosses the \(x\)-axis. For \(f(x) = \frac{3x}{(x+1)(x-2)}\), setting \(3x = 0\) is straightforward:- Solving gives \(x = 0\).Thus, the graph crosses the \(x\)-axis at \((0, 0)\). This point becomes crucial when sketching the graph as it shows where the function changes signs as it crosses the \(x\)-axis.

Sign Analysis

Sign analysis helps us understand the intervals where the function is positive or negative. For \(f(x) = \frac{3x}{(x+1)(x-2)}\), we test values in the intervals defined by vertical asymptotes and \(x\)-intercepts:

• For \((-\infty, -1)\), choose \(x = -2\), and find \(f(-2) > 0\).
• For \((-1, 0)\), choose \(x = -0.5\), and find \(f(-0.5) < 0\).
• For \((0, 2)\), choose \(x = 1\), and find \(f(1) < 0\).
• For \((2, \infty)\), choose \(x = 3\), and find \(f(3) > 0\).

This information is plotted around the asymptotes and \(x\)-intercepts, forming key parts of a function's graph. Each interval tells us whether the graph lies above or below the \(x\)-axis, significantly aiding in accurately sketching the curve.