Problem 40 - 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. $$f(x)=\frac{(x+3)(x-5)}{(x+1)(x-4)}$$

Step-by-Step Solution

Answer

The graph has x-intercepts at (-3, 0) and (5, 0), y-intercept at (0, 15/4), vertical asymptotes at x = -1 and x = 4, and horizontal asymptote at y = 1.

Step 1: Identify the x-intercepts

The x-intercepts of the rational function occur where the numerator equals zero and the denominator is not zero. Set the numerator \((x+3)(x-5)\) equal to zero.\[(x+3)(x-5) = 0\]Solve for \(x\):\(x+3=0\) or \(x-5=0\), which gives \(x = -3\) and \(x = 5\). So, the x-intercepts are \((x, y) = (-3, 0)\) and \((5, 0)\).

Step 2: Find the y-intercept

The y-intercept occurs when \(x = 0\). Substitute \(x = 0\) into the function and solve for \(f(0)\):\[f(0) = \frac{(0+3)(0-5)}{(0+1)(0-4)} = \frac{3 imes -5}{1 imes -4} = \frac{-15}{-4} = \frac{15}{4}\]Thus, the y-intercept is \(\left(0, \frac{15}{4}\right)\).

Step 3: Determine the vertical asymptotes

Vertical asymptotes occur where the denominator equals zero and the numerator is not zero. Set the denominator \((x+1)(x-4)\) equal to zero.\[(x+1)(x-4) = 0\]Solve for \(x\):\(x+1=0\) or \(x-4=0\), which gives \(x = -1\) and \(x = 4\). Therefore, there are vertical asymptotes at \(x = -1\) and \(x = 4\).

Step 4: Identify horizontal asymptotes

The horizontal asymptote can be determined by comparing the degrees of the numerator and the denominator. Both the numerator and the denominator are of degree 2 (highest power \(x^2\)).For rational functions where the degrees are the same, the horizontal asymptote is \(y = \frac{a}{b}\), where \(a\) and \(b\) are the leading coefficients. Here, both leading coefficients are 1, so the horizontal asymptote is \(y = 1\).

Step 5: Sketch the graph

With the information gathered:- Mark the x-intercepts at \((-3, 0)\) and \((5, 0)\).- Mark the y-intercept at \(\left(0, \frac{15}{4}\right)\).- Draw vertical asymptotes at \(x = -1\) and \(x = 4\).- Sketch the horizontal asymptote at \(y = 1\).- Use these points and asymptotes as guides to sketch the behavior of the graph approaching asymptotes and passing through intercepts.

Key Concepts

X-Intercepts

The x-intercepts of a rational function are crucial points where the graph crosses the x-axis. These occur when the numerator equals zero, but the denominator does not. To find x-intercepts for the function \( f(x) = \frac{(x+3)(x-5)}{(x+1)(x-4)} \), we set the numerator equal to zero:

• \((x+3) = 0\) leads to \(x = -3\)
• \((x-5) = 0\) leads to \(x = 5\)

In this case, the x-intercepts are at \((-3, 0)\) and \((5, 0)\). At these points, the graph touches or crosses the x-axis. Identifying these intercepts is the first step when sketching any rational function graph as they show where the graph impacts the horizontal axis.

Y-Intercepts

The y-intercept of a rational function is the point where the graph crosses the y-axis. Unlike x-intercepts, there is usually only one y-intercept for a function. You find the y-intercept by substituting \(x = 0\) into the function and solving for \(f(0)\). For our function, \[ f(0) = \frac{(0+3)(0-5)}{(0+1)(0-4)} = \frac{3 \times -5}{1 \times -4} = \frac{-15}{-4} = \frac{15}{4} \]Thus, the y-intercept is \( \left(0, \frac{15}{4} \right) \). This point indicates where the graph intersects the vertical axis and provides a starting point for analyzing how the function behaves as it moves away from the y-axis.

Vertical Asymptotes

Vertical asymptotes indicate where a rational function approaches infinity or negative infinity. These occur where the denominator equals zero, but not the numerator. Our function \( f(x) = \frac{(x+3)(x-5)}{(x+1)(x-4)} \) has a denominator of \((x+1)(x-4)\). Setting each factor of the denominator to zero:

• \(x + 1 = 0\) gives \(x = -1\)
• \(x - 4 = 0\) gives \(x = 4\)

Thus, vertical asymptotes are located at \(x = -1\) and \(x = 4\). As \(x\) approaches these values, the function values grow larger in magnitude, going towards positive or negative infinity. Graphically, vertical asymptotes represent lines that the graph never touches or crosses, indicating a division by zero.

Horizontal Asymptotes

Horizontal asymptotes give us insight into how a rational function behaves as \(x\) becomes very large or very small. They depend on the degrees of the numerator and the denominator. For \( f(x) = \frac{(x+3)(x-5)}{(x+1)(x-4)} \), both numerator and denominator are of degree 2.When the degrees are the same, the horizontal asymptote is the ratio of the leading coefficients of the numerator and the denominator. Here, both coefficients are 1, so the horizontal asymptote is \( y = 1 \). This means as \(x\) goes to positive or negative infinity, the function value approaches 1. Horizontal asymptotes help us predict the end behavior of the function, showing that while the graph may cross these lines, it eventually levels out.