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Chapter 0: Problem 21

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

Short Answer

Expert verified
The function is symmetric about the y-axis, no x-intercepts, y-intercept at (0, 1).

Step by step solution

01

Identify Type of Function

The equation given is \( y = \frac{1}{x^2 + 1} \). This is a rational function because it is the ratio of two polynomials. The denominator \( x^2 + 1 \) is always positive and never zero, so the function is defined for all real numbers.
02

Examine Symmetry

To check symmetry with respect to the y-axis, test \( f(-x) \): \( f(-x) = \frac{1}{(-x)^2 + 1} = \frac{1}{x^2 + 1} = f(x) \). Thus, the function is even and symmetric about the y-axis. There's no need to check for origin symmetry as it's not applicable for even functions.
03

Find x-intercepts

To find x-intercepts, set \( y = 0 \) and solve: \( \frac{1}{x^2 + 1} = 0 \). This equation has no solution because \( \frac{1}{x^2+1} \) never equals zero (as the smallest value it can be is \( \frac{1}{1} \)). Hence, there are no x-intercepts.
04

Find y-intercept

To find the y-intercept, set \( x = 0 \) in the equation: \( y = \frac{1}{0^2 + 1} = 1 \). Therefore, the y-intercept is (0, 1).
05

Plot Key Points and Sketch Graph

Evaluate the function for a few key values of \( x \) to sketch the graph. For example, at \( x = 1, y = \frac{1}{2} \) and \( x = 2, y = \frac{1}{5} \); similarly for \( x = -1 \) and \( x = -2 \). Due to symmetry, the graph will mirror across the y-axis. Plot these points and the y-intercept, then sketch the smooth curve approaching the x-axis but never touching it.

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Key Concepts

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

Symmetry in Graphs
When graphing functions, one of the key properties to check for is symmetry. This can simplify the graphing process, making it easier to identify points and sketch the curve. A function is symmetric with respect to the y-axis if substituting \(-x\) in place of \(x\) gives the original function back: \(f(-x) = f(x)\). This type of function is known as an even function, which means you will see a mirrored image of one half of the graph over the y-axis. In our exercise, the function \(y = \frac{1}{x^2 + 1}\) is an even function, showing symmetry about the y-axis.
Finding Intercepts
Intercepts are points where the graph crosses the axes. They help understand where the function meets the specified horizontal and vertical lines. The x-intercept occurs where the graph crosses the x-axis and the y-value is zero.
  • To find x-intercepts: Set \(y = 0\) and solve for \(x\).
  • To find y-intercepts: Set \(x = 0\) and solve for \(y\).
In the given function \(y = \frac{1}{x^2 + 1}\), there are no x-intercepts because the equation \(\frac{1}{x^2 + 1} = 0\) has no real solutions; the smallest value of \(x^2 + 1\) is 1, never reaching zero.However, the y-intercept is easily found by setting \(x = 0\) which yields \(y = 1\). Thus, the graph crosses the y-axis at the point (0, 1).
Rational Functions
Rational functions consist of a ratio of two polynomials. They have unique characteristics like asymptotic behavior, which means the graph approaches a line but never truly touches it. For example, the given function \( y = \frac{1}{x^2 + 1} \) has the polynomial \( x^2 + 1 \) in the denominator, which determines the domain and behavior of the function across its graph.
  • Key Properties: Rational functions may have asymptotes. Vertical asymptotes relate to values that make the denominator zero, while horizontal asymptotes concern the behavior as \(x\) trends towards infinity.
  • For \( y = \frac{1}{x^2 + 1} \), no vertical asymptote exists as \(x^2 + 1\) never equals zero.
Thus, the graph steadily approaches the x-axis horizontally, but never intersects or meets it.
Even Functions
Even functions are intriguing due to their symmetry about the y-axis. This characteristic arises when \(f(-x) = f(x)\), reflecting each point over the y-axis.
  • In practical terms, knowing a function is even implies one side of the y-axis mirrors the other. This can help in sketching the graph.
  • For \(y = \frac{1}{x^2 + 1}\), calculated symmetrically at \(x = -1\) and \(x = 1\), the y-values are identical, such as \(f(-1) = f(1)\)
Being aware that a function is even can simplify graphing and analysis tasks, as it reduces the work by focusing efforts on one side of the graph.

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Most popular questions from this chapter

Circular motion can be modeled by using the parametric representations of the form \(x(t)=\sin t\) and \(y(t)=\cos t\) (A parametric representation means that a variable, \(t\) in this case, determines both \(x(t)\) and \(y(t) .\) This will give the full circle for \(0 \leq t \leq 2 \pi .\) If we consider a 4 -foot-diameter wheel making one complete rotation clockwise once every 10 seconds, show that the motion of a point on the rim of the wheel can be represented by \(x(t)=2 \sin (\pi t / 5)\) and \(y(t)=2 \cos (\pi t / 5)\) (a) Find the positions of the point on the rim of the wheel when \(t=2\) seconds, 6 seconds, and 10 seconds. Where was this point when the wheel started to rotate at \(t=0 ?\) (b) How will the formulas giving the motion of the point change if the wheel is rotating counterclockwise. (c) At what value of \(t\) is the point at (2,0) for the first time?

Write \(p(x)=1 / \sqrt{x^{2}+1}\) as a composite of four functions.

Plot the graph of each equation. Begin by checking for symmetries and be sure to find all \(x\) - and \(y\) -intercepts. \(x^{2}+y^{2}=4\)

Plot the graph of each equation. Begin by checking for symmetries and be sure to find all \(x\) - and \(y\) -intercepts. \(2 x^{2}-4 x+3 y^{2}+12 y=-2\)

Suppose that a continuous function is periodic with period 2 and is quadratic between -0.25 and 0.25 and linear between -1.75 and \(-0.25 .\) In addition, it has the value 0 at 0 and 0.0625 at \(\pm 0.25 .\) Sketch the function over the domain \([-2,2],\) and give a piecewise definition of the function.
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