Problem 19 - Precalculus: Functions and Graphs

Question

Sketch the graph of \(f\) $$f(x)=\frac{x-3}{x^{2}-1}$$

Step-by-Step Solution

Answer

Domain excludes \(x = 1\) and \(x = -1\); vertical asymptotes at \(x = 1\) and \(x = -1\); horizontal asymptote at \(y = 0\); x-intercept at \(x = 3\). Graph approaches these asymptotes accordingly.

Step 1: Identify the Domain

The function is not defined when the denominator equals zero. So, we find when \( x^2 - 1 = 0\). This factors into \((x-1)(x+1) = 0\), giving us \(x = 1\) and \(x = -1\). Therefore, the domain is all real numbers except \(x = 1\) and \(x = -1\).

Step 2: Determine the Vertical Asymptotes

Vertical asymptotes occur at values that make the denominator zero but not the numerator. From the factorization \((x-1)(x+1) = 0\), we have vertical asymptotes at \(x = 1\) and \(x = -1\).

Step 3: Find the Horizontal Asymptote

To find the horizontal asymptote, we consider the behavior as \(x\) approaches infinity or negative infinity. Divide the leading coefficients: \(\frac{1}{x}\approx 0\), so the horizontal asymptote is \(y = 0\).

Step 4: Find the x-intercept

The x-intercept occurs where \(f(x) = 0\). Set the numerator equal to zero: \(x - 3 = 0\), so the x-intercept is at \(x = 3\).

Step 5: Analyze the Sign Change

Determine the sign of \(f(x)\) in the intervals marked by the intercepts and asymptotes. Compute \(f(x)\) for test points in the intervals \((-\infty, -1)\), \((-1, 1)\), \((1, 3)\), and \((3, \infty)\) to see where \(f(x)\) is positive or negative.

Step 6: Draw the Graph

Sketch the graph using the asymptotes, x-intercept, and test point evaluations. The graph approaches the horizontal asymptote \(y=0\) as \(x\to\pm\infty\), and near the vertical asymptotes at \(x=1\) and \(x=-1\), the graph will rise to \(\pm \infty\) or fall to \(\mp \infty\).

Key Concepts

Domain of a Function

The domain of a function is the set of all possible input values (typically represented as \(x\)-values) that will result in a real number for the output. In the context of rational functions, which are ratios of polynomials, it is crucial to identify where the denominator equals zero, as division by zero is undefined. For the function \(f(x) = \frac{x-3}{x^2-1}\), this means identifying when the denominator \(x^2 - 1 = 0\).

• This equation can be factored into \((x-1)(x+1) = 0\).
• Setting each factor equal to zero gives \(x = 1\) and \(x = -1\).

Thus, the domain of \(f(x)\) is all real numbers except \(x = 1\) and \(x = -1\). You can write the domain in interval notation as \((-\infty, -1) \cup (-1, 1) \cup (1, \infty)\). Remember, identifying the domain helps in understanding where the function is defined and continuous.

Vertical Asymptotes

Vertical asymptotes are vertical lines that the graph of a function approaches but never actually touches. These occur in rational functions at values of \(x\) that make the denominator zero, provided those values do not also make the numerator zero (in which case, you'd likely have a hole instead). For our function \(f(x) = \frac{x-3}{x^2-1}\), we determined the problematic points for the domain as \(x=1\) and \(x=-1\) by solving \(x^2 - 1 = 0\).

• Since substituting \(x = 1\) or \(x = -1\) does not zero out the numerator \(x - 3\), these are the locations of the vertical asymptotes.
• Graphically, as \(x\) approaches \(1\) or \(-1\), the function values skyrocket toward infinity or plummet to negative infinity.

Vertical asymptotes indicate a type of discontinuity where the function is undefined and separate from its behavior on either side of these \(x\)-values.

Horizontal Asymptotes

Horizontal asymptotes describe the behavior of a graph as \(x\) moves towards positive or negative infinity. They are horizontal lines that the graph approaches but does not necessarily cross or reach. For rational functions, the horizontal asymptote is often a simple function of the degrees of the polynomials in the numerator and denominator. For \(f(x) = \frac{x-3}{x^2-1}\), we check the degrees:

• The numerator has a degree of 1.
• The denominator has a degree of 2.

When the degree of the denominator exceeds the degree of the numerator, the horizontal asymptote is \(y = 0\). This reflects that as \(|x|\) becomes very large, the effect of the numerator diminishes, and the fraction approaches zero. The horizontal asymptote indicates long-term behavior and helps with visualizing the end-behavior of the function's graph.

X-intercepts

The \(x\)-intercepts of a graph are the points where the graph intersects the \(x\)-axis. For rational functions, these are found by setting the numerator equal to zero (since the entire function will be zero when the numerator is zero, provided the denominator is not zero at the same point). For the function \(f(x) = \frac{x-3}{x^2-1}\), we solve for the \(x\)-intercept by setting the numerator, \(x-3\), to zero.

• This gives \(x-3 = 0\), which solves to \(x = 3\).

Hence, the \(x\)-intercept is at \(x=3\). This point shows where the graph crosses the \(x\)-axis, providing valuable information about the function's zeros and helping to guide the sketching of its graph.