Problem 31 - Algebra and Trigonometry Real Mathematics, Real People

Question

Sketch the graph of the rational function by hand. As sketching aids, check for intercepts, vertical asymptotes, horizontal asymptotes, and holes. Use a graphing utility to verify your graph. $$f(x)=\frac{x^{2}-1}{x+1}$$

Step-by-Step Solution

Answer

The function has the x-intercepts at x= -1 and x=1, y-intercept at y=-1, vertical asymptote at x=-1, and horizontal asymptote at y=1. There is a hole at \( x=-1 \) indicated by the cancelled factors. The graphing verification reveals this sketch to be accurate.

Step 1: Finding the Intercepts

To find the y-intercept, set \(x=0\) in the function. For the x-intercept, set the function equal to 0 and solve for x. Therefore, \(f(0)=\frac{0^{2}-1}{0+1}=-1\) gives us the y-intercept and setting \(f(x)=0\) gives us \(x²-1=0\), which when solved gives \(x=\pm1\) which are the two x-intercepts.

Step 2: Finding the Vertical Asymptotes

The vertical asymptotes occur where the denominator is zero and the numerator is non-zero. Therefore, you set the denominator equal to zero and solve: \(x+1=0\). Solving this yields \(x=-1\).

Step 3: Finding the Horizontal Asymptotes

Horizontal asymptotes are based on the degree of the polynomial in the numerator and denominator. Here, since the degree of the numerator is equal to the denominator, the horizontal asymptote is the ratio of the leading coefficients, which is \(y=1\).

Step 4: Finding Holes

A hole in the graph of a rational function occurs when a factor in the denominator cancels out with a factor in the numerator. Here factors \(x-1\) and \(x+1\) cancel out for \(x=-1\) thus a hole exists at \(x=-1\).

Step 5: Sketching the Graph

Now, combine all of this information to sketch the graph. Place the x and y-intercepts, plot the asymptotes, and mark the hole. To fill in the rest of the graph, choose a few points on either side of the vertical asymptote to calculate and plot.

Step 6: Verifying using a graphing utility

You should now verify your graph using a graphing utility. Input the function into the graphing software and compare the resulting graph with your sketch. They should match.

Key Concepts

Intercepts

Intercepts are key points where the graph of a function crosses the x-axis or y-axis. For rational functions, finding intercepts involves simple substitutions:

- **Y-Intercept**: This is where the graph crosses the y-axis, and we find it by setting the value of x to zero. In the given rational function, \( f(x) = \frac{x^2-1}{x+1} \), setting \( x = 0 \) gives \( f(0) = -1 \). Therefore, the y-intercept is at point (0, -1).

- **X-Intercepts**: These are points where the graph crosses the x-axis. We find them by setting the output, \( f(x) \), to zero and solving for x: \( x^2-1=0 \). Solving this, we find \( x = \pm 1 \). Hence, the x-intercepts are at the points (1, 0) and (-1, 0). These intercepts give us valuable data points for sketching the graph.

Vertical Asymptotes

Vertical asymptotes occur in a rational function where the denominator equals zero, and the function does not have a real solution. This situation creates a vertical line which the graph approaches but never touches or crosses.

In the function \( f(x) = \frac{x^2-1}{x+1} \), set the denominator \( x+1 \) to zero. Solving \( x+1=0 \) gives \( x=-1 \). This tells us that there is a vertical asymptote at \( x=-1 \).

The presence of a vertical asymptote divides the graph into distinct sections that approach this line either infinitely upwards or downwards. When sketching, make sure to include this in the drawing, showing how the graph behaves as it approaches \( x=-1 \).

Horizontal Asymptotes

Horizontal asymptotes help us predict the end behavior of a rational function as the input (x) approaches positive or negative infinity. Determining horizontal asymptotes involves comparing the degrees of the polynomials in the numerator and denominator.

In this case, both the numerator \( x^2-1 \) and the denominator \( x+1 \) have the same degree (the degree is 1). Therefore, the horizontal asymptote is given by the ratio of their leading coefficients. In the function \( f(x) = \frac{x^2-1}{x+1} \), the leading coefficients are both 1. Thus, the horizontal asymptote is \( y=1 \).

This means as \( x \) becomes very large or very small, the function approaches the line \( y=1 \). When drawing your graph, make sure to show this line to indicate how the function trends towards this value.

Holes in Rational Functions

A hole in the graph of a rational function happens when a factor in the numerator is canceled by a factor in the denominator. This results in an undefined point or a 'hole' at a specific x-value.

For the function \( f(x) = \frac{x^2-1}{x+1} \), after factoring, we have \( \frac{(x-1)(x+1)}{x+1} \). Here, \( x+1 \) in both the numerator and denominator cancel out, indicating a hole at \( x=-1 \).

Even though there is a vertical asymptote at \( x=-1 \), the canceled factor suggests a removable discontinuity, or hole, at this point. While sketching, mark this hole with an open circle on the graph where the function is not defined.