Problem 4 - A Graphical Approach to Precalculus with Limits

Question

Match the rational function in Column I with the appropriate description in Column II. Choices in Column II can be used only once. Do not use a calculator. \(\mathbf{I}\) \(f(x)=\frac{-3+x^{2}}{x^{2}}\) \(\mathbf{II}\) A. The \(x\) -intercept is \((-3,0)\) B. The \(y\) -intercept is \((0,5)\) C. The horizontal asymptote is \(y=4\) D. The vertical asymptote is \(x=-1\) E. There is a hole in its graph at \(x=-4\) F. The graph has an oblique asymptote. G. The \(x\) -axis is its horizontal asymptote. H. The \(y\) -axis is its vertical asymptote.

Step-by-Step Solution

Answer

The function matches description H: The y-axis is its vertical asymptote.

Step 1: Analyze the Rational Function

Given the function \( f(x) = \frac{-3 + x^2}{x^2} \), let's simplify it by dividing each term in the numerator by the denominator \( x^2 \). This simplifies to \( f(x) = \frac{x^2}{x^2} - \frac{3}{x^2} = 1 - \frac{3}{x^2} \).

Step 2: Determine Asymptotes

For the horizontal asymptote, consider the behavior of the function as \( x \to \infty \): \( f(x) \approx 1 \). Hence, the horizontal asymptote is \( y = 1 \), matching none of the options regarding horizontal asymptotes directly. However, we note the form implies the graph approaches the x-axis, so \( y = 0 \) is close to option G.For vertical asymptotes, the equation \( x^2 = 0 \) gives a vertical asymptote at \( x = 0 \) (matching option H), since \( x^2 \equiv 0 \) leads to division by zero.

Step 3: Check for Holes

Holes occur when both the numerator and denominator have a common factor that cancels out upon simplification. In this case, \( -3 + x^2 \) does not factor out with the denominator \( x^2 \), indicating no holes in the graph.

Step 4: Analyze Intercepts

The x-intercept occurs where the numerator is zero: \( -3 + x^2 = 0 \Rightarrow x = \sqrt{3} \) or \( x = -\sqrt{3} \). However, \((-3,0)\) is not an intercept, this eliminates option A.The y-intercept, found by evaluating \( f(0) \), does not exist due to the division by zero at \( x = 0 \), eliminating option B as well.

Step 5: Evaluate Remaining Descriptions

Option F requires an oblique asymptote, which occurs when the degree of the numerator exceeds the denominator by 1, not applicable here. No oblique asymptote is present in our function, eliminating option F.

Key Concepts

Horizontal Asymptotes

In rational functions, horizontal asymptotes indicate the end-behavior of a graph as the input values grow exceedingly large or small. They act as an invisible boundary that the function approaches but never quite reaches.
This concept is especially important as it shows how the function behaves as it heads towards positive or negative infinity. A horizontal asymptote can be found by comparing the degrees of the polynomial function in the numerator and the denominator. If these degrees are equal, the horizontal asymptote is the ratio of the leading coefficients of the numerator and the denominator.
For the function given in the exercise, the horizontal asymptote is determined after simplification to be 1, as the function becomes \(f(x) = 1 - \frac{3}{x^2}\). As \(x\) approaches infinity, \(\frac{3}{x^2}\) tends to zero, leaving \(f(x) \approx 1\). This matches the behavior described in the original problem setup, although the exercise points to how the x-axis often acts as the horizontal line for approximation.

Vertical Asymptotes

Vertical asymptotes occur in the graph of a rational function where the denominator equals zero but the numerator does not equal zero at those points. This results in values where the graph shoots off toward positive or negative infinity, marked as specific x-values which the function will never touch.
In the simplified function \(f(x) = 1 - \frac{3}{x^2}\), a vertical asymptote appears where \(x^2 = 0\), which simplifies to \(x = 0\). Unlike some functions where multiple x-values could be vertical asymptotes, in this instance, the vertical asymptote is clearly at \(x = 0\) since \(x^2\) is the only expression in the denominator.
The textbook problem ties this to option H where the y-axis becomes the vertical asymptote, reflecting a common scenario for many rational functions.

Intercepts

Intercepts describe the points where the function graph crosses the x-axis or the y-axis. For rational functions, intercepts can be found by setting the numerator equal to zero for x-intercepts and substituting zero for \(x\) in finding the y-intercept.
In the case of the rational function \(f(x) = \frac{-3 + x^2}{x^2}\), we identify x-intercepts where \(-3 + x^2 = 0\). Solving this results in \(x = \pm \sqrt{3}\), showing two x-intercepts at \((-\sqrt{3}, 0)\) and \((\sqrt{3}, 0)\). This eliminates the option A from the list as neither x-intercept matches (-3,0).
Finding a y-intercept involves setting \(x = 0\), but in this function, doing so results in division by zero, indicating a y-intercept does not exist. This aligns with the elimination of option B as the function does not cross or meet the y-axis at any point.

Graphing Rational Functions

Graphing rational functions requires understanding and combining the analysis of asymptotes and intercepts. By considering these factors, you’ll visualize how the graph behaves across its domain.
For our example \(f(x) = 1 - \frac{3}{x^2}\), we crafted an image based on:

• Horizontal asymptotes at \(y = 1\)
• Vertical asymptotes at \(x = 0\)
• X-intercepts at \((-\sqrt{3}, 0)\) and \((\sqrt{3}, 0)\).

When graphing, start by plotting the asymptotes as dashed lines to show the restrictions and expected behavior, then place intercepts. This particular function lacks a y-intercept but creates a distinctive shape, moving towards the horizontal line and avoiding the vertical one.
This process illustrates why correctly identifying intercepts and understanding asymptotic behavior is essential when picturing the complete rational function graph.