Problem 51 - College Algebra and Calculus: An Applied Approach

Question

Sketch the graph of the rational function. To aid in sketching the graphs, check for intercepts, symmetry, vertical asymptotes, and horizontal asymptotes. $$g(s)=\frac{s}{s^{2}+1}$$

Step-by-Step Solution

Answer

The graph of the rational function \( g(s) = \frac{s}{s^{2}+1} \) has origin symmetry, intercepts at (0,0), no vertical asymptotes, and a horizontal asymptote at \( y = 0 \).

Step 1: Find the Intercepts

Set \(g(s) = 0\). We get \(s = 0\). Thus, the y-intercept and x-intercept is at (0,0).

Step 2: Check for Symmetry

Replacing \(s\) with \(-s\) results in \(-g(s)\). Thus, the graph is odd and has origin symmetry.

Step 3: Find the Vertical Asymptotes

Set the denominator equal to zero and solve for \(s\). However, the equation \(s^2 + 1 = 0\) has no real solution. Therefore, there are no vertical asymptotes.

Step 4: Find the Horizontal Asymptotes

Determine the degree of the numerator and denominator. They are 1 and 2 respectively. A lower degree for the numerator indicates a horizontal asymptote at \(y = 0\)

Step 5: Graphing the Function

Using these points and asymptotes, sketch a graph that fits: a curve originating from the origin (because of the intercept at (0,0) and the function's origin symmetry) and approaching \(y = 0\) for large positive and negative \(s\) values (because of the horizontal asymptote at \(y = 0\)).

Key Concepts

Intercepts

Intercepts on a graph are the points where the function crosses the axes. For rational functions, these points are crucial as they offer starting points for sketching the graph. In our exercise, we find the intercepts by setting the function equal to zero. For the function \( g(s) = \frac{s}{s^2 + 1} \), setting \( g(s) = 0 \) gives us \( s = 0 \).
Hence, both the x-intercept and y-intercept are at the origin, (0,0). This indicates that the graph will pass through this point. By starting our sketch here, it becomes easier to understand how the function behaves around this central point.

Symmetry

Symmetry helps us understand the general shape of the graph and can simplify the graphing process. For a function to have symmetry, it usually reflects over the y-axis, the x-axis, or the origin. To check for symmetry in the function \( g(s) = \frac{s}{s^2 + 1} \), we substitute \( s \) with \(-s\) to see how the graph behaves.
This calculation shows that \( g(-s) = -g(s) \), indicating the function is odd. Odd functions have origin symmetry, meaning if you rotate the graph 180 degrees around the origin, it looks the same. Therefore, understanding this symmetry helps predict how the graph behaves across all quadrants.

Vertical Asymptotes

Vertical asymptotes are vertical lines that the graph approaches but never crosses. They occur at values of \( s \) that make the denominator zero, but not the numerator, signaling points where the function is undefined. For \( g(s) = \frac{s}{s^2 + 1} \), we check the denominator.
The equation \( s^2 + 1 = 0 \) has no real solutions as it results in imaginary values (since \( s^2 = -1 \) is not possible with real numbers). Hence, there are no vertical asymptotes for this function, indicating that the graph will not have breaks or division lines in real-numbered \( s \).

Horizontal Asymptotes

Horizontal asymptotes indicate the value that \( g(s) \) will approach as \( s \) becomes very large or very small. To find them, we compare the degrees of the polynomial in the numerator and the denominator. For \( g(s) = \frac{s}{s^2 + 1} \), the degree of the numerator is 1 (since \( s^1 \)) and the degree of the denominator is 2 (since \( s^2 \)).
When the degree of the numerator is less than the denominator, as here, the horizontal asymptote is at \( y = 0 \). This suggests that as \( s \) goes to positive or negative infinity, \( g(s) \) approaches zero but never quite reaches it. Recognizing this helps us sketch the graph's behavior at the extremes, confirming how the graph flattens out as it extends.