Problem 55 - Algebra and Trigonometry

Question

Find the intercepts and asymptotes, and then sketch a graph of the rational function. Use a graphing device to confirm your answer. \(s(x)=\frac{x^{2}-2 x+1}{x^{3}-3 x^{2}}\)

Step-by-Step Solution

Answer

The function's x-intercept is x = 1; vertical asymptotes are x = 0 and x = 3; horizontal asymptote is y = 0.

Step 1: Determine X-Intercepts

To find the x-intercepts of the function, set the numerator equal to zero and solve for x. The numerator is: \[ x^2 - 2x + 1 = 0 \]Factor the numerator:\[ (x - 1)^2 = 0 \]This implies x = 1 is the x-intercept.

Step 2: Determine Y-Intercept

To find the y-intercept, substitute x = 0 into the function and solve for s(x). \[ s(0) = \frac{0^2 - 2 \cdot 0 + 1}{0^3 - 3 \cdot 0^2} = \frac{1}{0} \]Since the denominator equals zero, there is no y-intercept because the function is not defined at x = 0.

Step 3: Find Vertical Asymptotes

Vertical asymptotes are found where the denominator is zero but the numerator is not zero. Denominator: \[ x^3 - 3x^2 = x^2(x - 3) = 0 \] Solutions are x = 0 and x = 3. Check the numerator at these points:For x = 0, numerator is 1 (not zero). Hence, vertical asymptote at x = 0.At x = 3, substitute into the numerator:\[ 3^2 - 2 \cdot 3 + 1 = 9 - 6 + 1 = 4 \] Hence, vertical asymptote at x = 3.

Step 4: Find Horizontal Asymptotes

For horizontal asymptotes, compare the degrees of the numerator and the denominator. The numerator has degree 2, and the denominator has degree 3. Since the numerator's degree is less than the denominator's degree, the horizontal asymptote is y = 0.

Step 5: Sketch the Graph

Utilize the x-intercept, vertical asymptotes, and the horizontal asymptote to sketch the graph. The function has an x-intercept at x = 1, vertical asymptotes at x = 0 and x = 3, and a horizontal asymptote at y = 0.

Step 6: Confirm Using Graphing Device

Use a graphing calculator or software to sketch the graph of \( s(x) = \frac{x^2 - 2x + 1}{x^3 - 3x^2} \), confirming the intercepts and asymptotes are accurate. Check if the graph behaves as expected near these points.

Key Concepts

Intercepts

When graphing a rational function, identifying the intercepts is an essential step. For our function \(s(x)=\frac{x^{2}-2 x+1}{x^{3}-3 x^{2}}\), we start by finding the intercepts.
**X-Intercepts:** To find where the graph crosses the x-axis, we set the numerator equal to zero and solve for \(x\). In this example, the numerator factors to \((x - 1)^2\), which simplifies to \(x = 1\). Thus, the graph intersects the x-axis at \(x=1\).
**Y-Intercepts:** Usually, the y-intercept is found by substituting \(x=0\) into the function. For this function, this leads to a division by zero, indicating no y-intercept. This occurs because the function is not defined at \(x=0\), often due to a vertical asymptote or hole at that point.

Asymptotes

Asymptotes provide critical insights into the behavior of rational functions. They're lines the graph approaches but never touches.
**Vertical Asymptotes** occur where the denominator is zero, but the numerator is not. For \(s(x)\), the denominator \(x^2(x - 3)\) is zero when \(x = 0\) and \(x = 3\). Checking the numerator at these points, we find values of 1 and 4, respectively, confirming vertical asymptotes at \(x=0\) and \(x=3\). The graph will tend toward infinity at these axes.
**Horizontal Asymptotes** show the end behavior of the function. We determine them by comparing the degrees of the numerator and the denominator. Here, the numerator has a degree of 2, and the denominator a degree of 3. Since the numerator's degree is lower, the horizontal asymptote is \(y=0\). This suggests that as \(x\) becomes very large or very small, the function approaches zero.

Graphing Calculator

A graphing calculator is an invaluable tool to visualize rational functions like \(s(x)=\frac{x^{2}-2 x+1}{x^{3}-3 x^{2}}\). When plotting this function:
- Enter the equation into the calculator to see the overall shape of the graph.
- Identify the intercepts, vertical asymptotes at \(x=0\) and \(x=3\), and the horizontal asymptote at \(y=0\).
The graphing calculator confirms these theoretical findings by showing the function's behavior near these critical points. As the function approaches \(x=0\) and \(x=3\), the graph tends to rise or fall sharply towards positive or negative infinity, depicting the vertical asymptotes.
Likewise, the graph flattens out along the x-axis, validating the horizontal asymptote at \(y=0\). Using such tools not only corroborates manual solutions but also enhances comprehension by visually demonstrating the broad characteristics of the function.