/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 54 Sketch the graph of the equation... [FREE SOLUTION] | 91影视

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Chapter 0: Problem 54

Sketch the graph of the equation. Identify any intercepts and test for symmetry. $$y=\frac{10}{x^{2}+1}$$

Short Answer

Expert verified
The graph of the function \(y = \frac{10}{x^2+1}\) will have no x-intercepts, a y-intercept at y = 10, and is symmetric around the y-axis.

Step by step solution

01

Plotting the Graph

To start with the graphing of the given function, plot some key points by choosing a few different x-values and corresponding y-values obtained from the function, \(y = \frac{10}{x^2+1}\). This will create an idea of how the actual graph looks.
02

Identifying the Intercepts

The x-intercept is found when the graph crosses the x-axis, where y = 0. Substituting y=0 in \(y = \frac{10}{x^2+1}\) will yield no solution, as the equation 鈭儀鈮�0 for y=0 cannot be satisfied. Therefore, the graph has no x-intercepts.\n\nThe y-intercept is found when the graph crosses the y-axis where x = 0. Substituting x=0 in \(y = \frac{10}{x^2+1}\) gives \(y = \frac{10}{(0)^2+1} = 10\). Therefore, the graph has a y-intercept at y = 10.
03

Testing for Symmetry

To test the equation for symmetry, it's needed to check the following conditions: \nIf the equation remains same upon replacing x with -x, it is symmetric about the y-axis. \nIf the equation remains same upon replacing y with -y, it's symmetric about the x-axis. \n\nReplacing x in the equation \(y = \frac{10}{x^2+1}\) with -x gives \(y = \frac{10}{(-x)^2+1} = \frac{10}{x^2+1}\), which means it's symmetric around the y-axis.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

X-Intercepts
Understanding the concept of x-intercepts is crucial for graphing rational functions. An x-intercept is a point where the graph of a function crosses the x-axis. To find the x-intercepts algebraically, we set the output value, y, to zero and solve for x. For the function in our exercise,

\[y = \frac{10}{x^2+1}\],
setting y to 0 does not provide any real numbers that satisfy the equation. This indicates that the graph never touches the x-axis and, therefore, has no x-intercepts. When graphing, this helps us understand that the function will either remain above or below the x-axis, depending on the sign of the numerator.
Y-Intercepts
The y-intercept of a graph is where it crosses the y-axis, which is when x equals zero. In our case, we substitute x with 0 in the function

\[y = \frac{10}{x^2+1}\],
which simplifies to \(y = 10\). Consequently, the graph has a single y-intercept at the point (0, 10). When graphing rational functions, identifying the y-intercept provides a pivotal starting point on the graph, aiding in the visualization of the function's behavior near the y-axis.
Symmetry Tests
Symmetry in functions can greatly simplify the graphing process. For rational functions, we commonly check for two types of symmetry: about the y-axis and the x-axis. For symmetry about the y-axis, we replace x with -x and see if the function remains unchanged.

Applying this to our function, \[y = \frac{10}{(-x)^2+1}\], yields the same formula because the square of a negative number equals the square of a positive number. Therefore, we conclude that the graph is symmetric about the y-axis.
Knowing this symmetry allows us to plot only half of the graph and replicate the other side across the y-axis, which simplifies our work.
Plotting Graphs
The process of plotting a graph for rational functions involves determining key features like intercepts and symmetry, as we've discussed, and then placing sufficient points to reveal the shape of the graph. Start with the intercepts and symmetry, then select a range of x-values and compute corresponding y-values to plot.

Since our function has no x-intercepts and is symmetric about the y-axis with a y-intercept at (0, 10), we can plot additional points for positive x-values and reflect them across the y-axis. Remember to note any vertical or horizontal asymptotes: in this case, the denominator suggests a horizontal asymptote at y = 0, since as x becomes very large, \(\frac{10}{x^2+1}\) approaches zero. When plotted, the graph will resemble a bell shape, peaking at the y-intercept and tending toward the asymptote without ever touching the x-axis.

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