Question

, plot the graph of each equation. Begin by checking for symmetries and be sure to find all \(x\) - and \(y\) -intercepts.. $$ y=\frac{x}{x^{2}+1} $$

Short answer

The function is odd with intercept (0,0), and horizontal asymptote y=0.

Step-by-step solution

Identify the Symmetries

To determine if the function has any symmetries, examine the function for evenness or oddness. The function is even if \(f(-x) = f(x)\) and odd if \(f(-x) = -f(x)\). For \(y = \frac{x}{x^2 + 1}\), substitute \(-x\) to get \(y = \frac{-x}{(-x)^2 + 1}\). Simplifying, we get \(y = -\frac{x}{x^2 + 1}\), which is \(-f(x)\), showing that the function is odd, thus symmetric about the origin.

Find the Intercepts

To find the \(x\)-intercepts, set \(y = 0\) and solve \(\frac{x}{x^2 + 1} = 0\). The solution is \(x = 0\).To find the \(y\)-intercept, substitute \(x = 0\) into the function: \(y = \frac{0}{0^2 + 1} = 0\). Thus, the graph intercepts both axes at the origin, \((0, 0)\).

Analyze the Behavior as \(x\to\pm\infty\)

As \(x\to\pm\infty\), the \(x^2\) term in the denominator dominates, so \(y = \frac{x}{x^2 + 1}\approx\frac{x}{x^2}=\frac{1}{x}\). Thus, as \(x\to\pm\infty\), \(y\to0\). The horizontal asymptote is \(y = 0\) (the x-axis).

Plot the Graph

Using the information gathered, plot the function \(y = \frac{x}{x^2 + 1}\). Start by marking the intercept point at the origin \((0,0)\). Note the symmetry about the origin and the horizontal asymptote at \(y=0\). Draw the curve passing through the origin, bending downwards towards the x-axis for both large positive and negative values of \(x\), reflecting the odd symmetric nature.

Key concepts

Symmetry in Functions

In mathematics, symmetry refers to a balanced and proportionate similarity found in two halves of an object, that is, one-half is the mirror image of the other half. Symmetries in functions are identified by how they behave when we substitute negative values and transform them.

When analyzing rational functions such as \( y = \frac{x}{x^2 + 1} \), we check for symmetry by investigating if the function is "even" or "odd".

• An even function satisfies \( f(-x) = f(x) \), being symmetric with respect to the y-axis.
• An odd function satisfies \( f(-x) = -f(x) \), indicating symmetry about the origin.

In our example, replacing \( x \) with \( -x \) results in \( \frac{-x}{x^2 + 1} = -f(x) \). Therefore, the graph of the function \( y = \frac{x}{x^2 + 1} \) is odd. It implies the graph will look the same if rotated 180 degrees around the origin. Such symmetry helps when sketching or predicting behavior on both sides of the plane.

Finding Intercepts

Intercepts are points where the graph intersects the axes. These points are crucial for understanding the shape and position of the graph.

Let's cover the two types you usually find:

• x-intercepts: Points where the graph crosses the x-axis. Here, \( y = 0 \). To find them, we set the equation \( \frac{x}{x^2 + 1} = 0 \). Solving gives \( x = 0 \), indicating that the graph touches the x-axis at the origin (0,0).
• y-intercepts: Points where the graph crosses the y-axis. Here, \( x = 0 \). Substitute \( x = 0 \) into the function: \( y = \frac{0}{0^2 + 1} = 0 \). So, the y-intercept is also at the origin (0,0).

In rational functions, intercepts lay the groundwork for plotting and provide intuitive understandings such as where the function values shift or start.

Horizontal Asymptotes

Horizontal asymptotes describe the end behavior of a function, particularly as \( x \) approaches infinity or negative infinity. While intercepts show where functions cross axes, asymptotes highlight the lines that graphs approach but typically never touch.

For the function \( y = \frac{x}{x^2 + 1} \), as \( x \to \pm \infty \), the term \( x^2 \) in the denominator becomes dominant. This means the expression behaves similarly to \( \frac{1}{x} \). As \( x \) becomes very large (positive or negative), \( \frac{1}{x} \) gets closer to zero. Therefore, we conclude:

• The horizontal asymptote is \( y = 0 \).

This insight indicates that no matter how large \( x \) becomes, the function \( y \) will tend towards zero, explaining why the graph hovers around the x-axis toward the ends.