Problem 50 - Algebra for College Students

Question

Find all asymptotes, \(x\) -intercepts, and \(y\) -intercepts for the graph of each rational function and sketch the graph of the function. $$f(x)=\frac{x}{x^{2}+1}$$

Step-by-Step Solution

Answer

Horizontal asymptote: . Both and -intercepts at (0,0). No vertical asymptotes.

Step 1: Finding the vertical asymptotes

Vertical asymptotes occur where the denominator equals zero, as long as the numerator does not also equal zero at those points. For the function , , , , , , , , , , , calculate the vertical asymptotes by setting the denominator equals to zero: , , , .

Step 2: Verifying vertical asymptotes

Solve for the vertical asymptotes: , . Since there are no real solutions for this equation, it means there are no vertical asymptotes for the given function.

Step 3: Finding horizontal asymptotes

For horizontal asymptotes of rational functions, compare the degrees of the numerator and the denominator. The degree of the numerator is 1, and the degree of the denominator is 2. When the degree of the numerator is less than the degree of the denominator, the horizontal asymptote is the x-axis or .

Step 4: Finding the y-intercepts

To find the -intercepts, set : . So the -intercept is at (0,0).

Step 5: Finding the x-intercepts

To find the -intercepts, set the numerator equal to zero and solve for : . So the -intercept is at (0,0).

Step 6: Sketching the graph

Using the information obtained, sketch the graph of the function. The graph will approach the horizontal asymptote as and will pass through the origin at (0,0), with no vertical asymptotes.

Key Concepts

Asymptotes

Asymptotes are lines that a graph approaches but never touches. They can be vertical, horizontal, or slant.

Vertical asymptotes occur where the denominator of a rational function equals zero, but only if the numerator doesn't also equal zero at that point. For the function \( f(x) = \frac{x}{x^2 + 1} \), we check if the denominator equals zero. However, solving \( x^2 + 1 = 0 \) gives us non-real solutions. This means there are no vertical asymptotes in this case.

Horizontal asymptotes depend on the degrees of the numerator and the denominator. For our function, the numerator is of degree 1 and the denominator is of degree 2. When the numerator's degree is less than the denominator's, the horizontal asymptote is \( y = 0 \). So, the horizontal asymptote here is the x-axis.

Intercepts

Intercepts are points where a graph crosses the x-axis or the y-axis.

To find the y-intercept, set \( x = 0 \) and solve for \( y \). For our function, if we substitute \( x = 0 \):

\[ f(0) = \frac{0}{0^2 + 1} = 0 \]

Thus, the y-intercept is at the point (0, 0).

Next, we find the x-intercept by setting the numerator equal to zero and solving for x. For \( f(x) = \frac{x}{x^2 + 1} \), this gives us:

\[ x = 0 \]

Thus, the x-intercept is also at the origin, (0, 0). So, both intercepts for our function are at (0, 0).

Graphing Rational Functions

Graphing rational functions involves plotting points and understanding the behavior near asymptotes and intercepts. We use the information from finding asymptotes and intercepts to sketch the graph.

Consider these steps:

• Locate vertical and horizontal asymptotes. For \( f(x) = \frac{x}{x^2 + 1} \), we have a horizontal asymptote at \( y = 0 \) and no vertical asymptotes.
• Plot the intercepts. The x-intercept and y-intercept are both at (0, 0).
• Determine the behavior of the function as \( x \) approaches large positive or negative values. Since the horizontal asymptote is \( y = 0 \), the function's values get closer to zero as \( x \) becomes very large or very small.
• Plot a few more points if needed for accuracy.

Combining this, the graph approaches the horizontal asymptote \( y = 0 \) as \( x \) tends towards infinity or minus infinity and will pass through the origin at (0, 0).

Remember, practice helps in getting a good grasp of how these elements work together in graphing.