Problem 94 - A Graphical Approach to Precalculus with Limits

Question

Use a calculator to graph rational function in the window indicated. Then (a) give the \(x\) - and y-intercepts, (b) explain why there are no vertical asymptotes, (c) give the equation of the oblique asymptote, and (d) give the domain and range. $$f(x)=\frac{4 x^{3}+8 x^{2}-36 x-72}{2 x^{2}-x+6} ;[-5,5] \text { by }[-20,15]$$

Step-by-Step Solution

Answer

The x-intercepts are \((3, 0)\) and \((-2, 0)\), y-intercept is \((0, -12)\), no vertical asymptotes, oblique asymptote is \(y = 2x + 5\), and domain/range are all real numbers.

Step 1: Graphing the Function

Use a graphing calculator and enter the rational function \( f(x) = \frac{4x^3 + 8x^2 - 36x - 72}{2x^2 - x + 6} \). Set the viewing window to \([-5, 5] \) for the x-values and \([-20, 15]\) for the y-values. Analyze the graph you see on the screen.

Step 2: Finding the Intercepts

To find the x-intercepts, set \(f(x) = 0\) which implies solving \(4x^3 + 8x^2 - 36x - 72 = 0\). Use a calculator to find that \(x = 3, 3, -2\). Thus, the x-intercepts are at \((3, 0)\) and \((-2, 0)\). For the y-intercept, evaluate \(f(0)\), which gives \(f(0) = \frac{-72}{6} = -12\). The y-intercept is \((0, -12)\).

Step 3: Analyzing Vertical Asymptotes

Vertical asymptotes occur where the denominator is zero and the numerator is not zero. Solve \(2x^2 - x + 6 = 0\) to check for real roots, but find that there are no real solutions, indicating no vertical asymptotes.

Step 4: Finding the Oblique Asymptote

Since the degree of the numerator is one higher than that of the denominator, an oblique asymptote exists. Use polynomial long division to divide \(4x^3 + 8x^2 - 36x - 72\) by \(2x^2 - x + 6\). This results in the quotient \(2x + 5\), so the oblique asymptote is \(y = 2x + 5\).

Step 5: Determining Domain and Range

The domain includes all real numbers, as there are no restrictions from the denominator (no zeros). From the graph, estimate the range. Since there are no vertical asymptotes but there is an oblique asymptote, the function approaches all y-values except possibly near the asymptote equations and intercepts. The range is approximately all real numbers \((-orall, orall)\).

Key Concepts

Graphing Calculators

Graphing calculators are powerful tools that help us visualize complex functions such as rational functions. To graph a rational function like \( f(x) = \frac{4x^3 + 8x^2 - 36x - 72}{2x^2 - x + 6} \), you'll first need to enter the function correctly into the calculator. Ensure that you input the whole expression with parentheses around the numerator and denominator.
Once entered, set your graphing window based on the exercise instructions. Here, use

• \([-5, 5]\) for the x-values,
• and \([-20, 15]\) for the y-values.

These settings provide enough space to display important features of the graph, such as intercepts and asymptotes, clearly. By examining the displayed graph, you will be able to understand how the rational function behaves over these intervals.

Intercepts

Intercepts are points where the graph of a function crosses the axes. For the rational function in question, we find these by setting the function equal to zero for x-intercepts and evaluating for \( x = 0 \) for the y-intercept.
**X-intercepts** are found by solving \( 4x^3 + 8x^2 - 36x - 72 = 0 \). When you solve this, you find the solutions \( x = 3, 3, -2 \), which shows that the x-intercepts are at \((3, 0)\) and \((-2, 0)\). These are the points where the graph touches the x-axis.
**Y-intercept** is calculated by finding \( f(0) \). In this case, \( f(0) = \frac{-72}{6} = -12 \), yielding a y-intercept at \((0, -12)\). This is where the graph crosses the y-axis. Identifying intercepts is crucial for understanding the overall shape and position of the graph.

Asymptotes

Asymptotes of a graph are lines that the graph approaches but never touches. They may be vertical, horizontal, or oblique, and they give us a lot of information about the behavior of a function.
**Vertical Asymptotes** usually occur where the denominator of a rational function is zero while the numerator is not zero. However, for this function, the equation \( 2x^2 - x + 6 = 0 \) has no real solutions, meaning there are no vertical asymptotes.
**Oblique Asymptotes** occur when the degree of the numerator is one higher than the degree of the denominator. We find it by performing polynomial long division. By dividing \( 4x^3 + 8x^2 - 36x - 72 \) by \( 2x^2 - x + 6 \), we find the oblique asymptote is \( y = 2x + 5 \). Thus, the graph will approach this line as we move towards infinity.

Domain and Range

The domain of a rational function consists of all possible x-values. In this exercise, since the denominator \( 2x^2 - x + 6 \) doesn't equate to zero anywhere within real numbers, the domain is all real numbers \( (-\infty, \infty) \).
The range involves all possible y-values the function can take. Due to the lack of vertical asymptotes but the presence of an oblique asymptote \( y = 2x + 5 \), the graph covers almost every y-value. The graph stretches to infinity in both directions, approximating the line but never actually reaching or crossing it except for intercept points.
Hence, the range is approximately all real numbers \((-\infty, \infty)\), excluding the values immediately adjacent to the oblique asymptote, which it approaches but never touches. Understanding the domain and range gives insight into the limitations and behavior of the function.