Problem 58 - Precalculus: Functions and Graphs

Question

\( Graph f,\) and find equations of the vertical asymptotes. $$f(x)=\frac{x^{2}-9.01}{x-3}$$

Step-by-Step Solution

Answer

The equation of the vertical asymptote is \(x = 3\).

Step 1: Identify the form of the function

The given function is a rational function in the form \[f(x) = \frac{n(x)}{d(x)}\]where the numerator \(n(x) = x^2 - 9.01\) and the denominator \(d(x) = x - 3\).

Step 2: Set the denominator to zero

To find the vertical asymptotes, we need to determine where the denominator of the function equals zero. Solve the equation\[x - 3 = 0\]to find the critical points that could lead to vertical asymptotes.

Step 3: Solve for x

Solve the equation from Step 2:\[x = 3\]This gives us a potential vertical asymptote at \(x = 3\).

Step 4: Verify the potential asymptote

Check if the numerator is also zero at \(x = 3\) to confirm a vertical asymptote (as opposed to a removable discontinuity). Substitute \(x = 3\) into the numerator:\[3^2 - 9.01 = 9 - 9.01 = -0.01\]Since the numerator is not zero, there is no removable discontinuity at \(x = 3\), confirming that \(x = 3\) is indeed a vertical asymptote.

Step 5: Graph the Function

To graph the function, plot points around \(x = 3\) and observe the behavior as \(x\) approaches 3 from the left and right. The graph will show an undefined point and the curve will approach infinity or negative infinity near \(x = 3\), highlighting the vertical asymptote.

Key Concepts

Vertical Asymptotes

A vertical asymptote in a rational function represents a line where the function approaches infinity or negative infinity. It visually appears as a vertical line on a graph. For the function \(f(x) = \frac{x^2 - 9.01}{x - 3}\), finding a vertical asymptote involves setting the denominator \(d(x)\) to zero. This gives us critical points where the function is undefined or may have asymptotic behavior.
In this specific exercise, we create the equation \(x - 3 = 0\). Solving this, we find \(x = 3\). Now, to confirm \(x = 3\) is indeed a vertical asymptote, it's important to check if substituting \(x = 3\) makes the numerator zero. Here, \(3^2 - 9.01 = -0.01\) isn't zero. Therefore, there is no removable discontinuity, and we can confidently state that the vertical asymptote occurs at \(x = 3\). This means as \(x\) gets close to 3, the function's value will shoot towards positive or negative infinity.

Removable Discontinuity

Removable discontinuities occur when the numerator and denominator both approach zero at the same point. This results in a 0/0 form, which is undefined mathematically but can be "removed" by canceling common factors. In our function \(f(x) = \frac{x^2 - 9.01}{x - 3}\), substituting \(x = 3\) into the numerator gives \(-0.01\), not zero. Hence, there is no 0/0 indeterminate form.

When such forms do occur, offering a removable discontinuity, it implies the equation can be simplified. However, this is not the case here. If \(x = 3\) made both the numerator and denominator zero, we could factor and simplify the expression to evaluate the rational limit as \(x\) approaches our point of interest. In simpler terms, removable discontinuities are fixable gaps in the graph where we'd normally be able to draw a continuous curve, unlike vertical asymptotes, which are fixed lines that cannot be crossed or smoothed.

Graphing Functions

Graphing rational functions involves plotting key points and observing behavior around undefined points like vertical asymptotes. For the function \(f(x) = \frac{x^2 - 9.01}{x - 3}\), the graph changes drastically as \(x\) approaches the vertical asymptote at 3.

To graph the function effectively:

• Select several points around the vertical asymptote, such as \(x = 2.9, 3, and 3.1\), to predict the graph's behavior.
• Notice that as \(x\) gets closer to 3, the function's value approaches infinity or negative infinity, confirming the asymptote visually.
• Plot additional points for a broader understanding of the function's curve away from the asymptote. This wider look ensures you capture the global behavior of the graph.

Overall, the graph will highlight how the function cannot pass through \(x = 3\) due to the asymptotic nature, with steep inclines or declines as \(x\) nears this critical point. Watching for these features is crucial in understanding the full scope of rational functions.