Problem 22
Question
In Problems 1-30, plot the graph of each equation. Begin by checking for symmetries and be sure to find all \(x\) - and \(y\)-intercepts. $$ y=\frac{x}{x^{2}+1} $$
Step-by-Step Solution
Verified Answer
The graph is symmetric about the origin, intercepts at \((0,0)\), and has a horizontal asymptote at \(y=0\).
1Step 1: Identify Symmetries
To determine symmetries, check for symmetry about the y-axis, x-axis, and origin. Replace \(x\) with \(-x\) in the equation to check for y-axis symmetry. The equation becomes \(y = \frac{-x}{x^2 + 1}\), which is not equal to the original, indicating no y-axis symmetry. For origin symmetry, replace \(x\) with \(-x\) and \(y\) with \(-y\), giving \(-y = \frac{-x}{x^2 + 1}\), which simplifies to the original equation: \(y = \frac{x}{x^2 + 1}\). Therefore, the graph has origin symmetry.
2Step 2: Find x-Intercepts
To find the x-intercepts, set \(y = 0\) and solve for \(x\). The equation \(0 = \frac{x}{x^2 + 1}\) implies \(x = 0\). Therefore, the x-intercept is at \((0, 0)\).
3Step 3: Find y-Intercepts
To find the y-intercepts, set \(x = 0\) and solve for \(y\). Substituting, we get \(y = \frac{0}{0^2 + 1} = 0\). Therefore, the y-intercept is at \((0, 0)\).
4Step 4: Analyze Asymptotic Behavior
Examine the behavior of \(y\) as \(x\) approaches infinity or negative infinity. For large \(|x|\), \(y = \frac{x}{x^2 + 1}\) approaches \(y \approx \frac{x}{x^2} = \frac{1}{x}\), indicating that \(y\) approaches 0 but never actually reaches it. Thus, the equation has a horizontal asymptote at \(y = 0\).
5Step 5: Sketch the Graph
Plot the points found: the intercept at \((0, 0)\), and the asymptotic behavior as \(x\) grows large in either direction. The function is odd and symmetric about the origin. As \(x\) moves away from zero, the value of \(y\) approaches but never reaches 0. The curve will reside primarily in the first and third quadrants, moving from the second to the fourth as \(x\) crosses zero.
Key Concepts
Symmetry in GraphsX-Intercepts and Y-InterceptsAsymptotic BehaviorSketching Graphs
Symmetry in Graphs
Symmetry in a graph helps us understand the overall shape and behavior of a function. For most graphs, we typically look for symmetry in three ways: around the y-axis, the x-axis, and the origin.
To check for y-axis symmetry, replace every instance of \(x\) in the equation with \(-x\). For our equation, this results in \(y = \frac{-x}{x^2 + 1}\). Since this does not match the original equation, there is no y-axis symmetry. Next, for origin symmetry, substitute \(x\) with \(-x\) and \(y\) with \(-y\), leading to \(-y = \frac{-x}{x^2 + 1}\). Simplifying this, we find that it results in the original equation, confirming origin symmetry.
This origin symmetry indicates that the graph is a mirror image of itself when rotated 180 degrees around the origin. Such symmetry is often seen in odd functions.
To check for y-axis symmetry, replace every instance of \(x\) in the equation with \(-x\). For our equation, this results in \(y = \frac{-x}{x^2 + 1}\). Since this does not match the original equation, there is no y-axis symmetry. Next, for origin symmetry, substitute \(x\) with \(-x\) and \(y\) with \(-y\), leading to \(-y = \frac{-x}{x^2 + 1}\). Simplifying this, we find that it results in the original equation, confirming origin symmetry.
This origin symmetry indicates that the graph is a mirror image of itself when rotated 180 degrees around the origin. Such symmetry is often seen in odd functions.
X-Intercepts and Y-Intercepts
Intercepts are points where the graph crosses the axes. They provide key information about where the function is zero (for x-intercepts) and the starting value of the graph (for y-intercepts).
For x-intercepts, we set \(y = 0\) and solve for \(x\). In the equation \(0 = \frac{x}{x^2 + 1}\), multiplying both sides by \(x^2 + 1\) gives \(x = 0\). Thus, the x-intercept is at the origin, or point \((0,0)\).
For y-intercepts, set \(x = 0\) and solve for \(y\). Plugging into the equation gives \(y = \frac{0}{0^2 + 1} = 0\), confirming the y-intercept is also at \((0,0)\).
The location of both intercepts at \((0,0)\) gives further insights into the steepness and orientation of the graph at the origin.
For x-intercepts, we set \(y = 0\) and solve for \(x\). In the equation \(0 = \frac{x}{x^2 + 1}\), multiplying both sides by \(x^2 + 1\) gives \(x = 0\). Thus, the x-intercept is at the origin, or point \((0,0)\).
For y-intercepts, set \(x = 0\) and solve for \(y\). Plugging into the equation gives \(y = \frac{0}{0^2 + 1} = 0\), confirming the y-intercept is also at \((0,0)\).
The location of both intercepts at \((0,0)\) gives further insights into the steepness and orientation of the graph at the origin.
Asymptotic Behavior
Asymptotic behavior describes how a function behaves as it approaches infinity or an undefined point. In simpler terms, it's about tracking where the function is headed without ever actually reaching there.
For the function \(y = \frac{x}{x^2 + 1}\), as \(x\) becomes very large (positive or negative), the term \(x^2 + 1\) dominates, and \(y\) approximates to \(\frac{x}{x^2} = \frac{1}{x}\).
As \(x\) approaches infinity, \(\frac{1}{x}\) approaches 0, signaling a horizontal asymptote at \(y = 0\).
For the function \(y = \frac{x}{x^2 + 1}\), as \(x\) becomes very large (positive or negative), the term \(x^2 + 1\) dominates, and \(y\) approximates to \(\frac{x}{x^2} = \frac{1}{x}\).
As \(x\) approaches infinity, \(\frac{1}{x}\) approaches 0, signaling a horizontal asymptote at \(y = 0\).
- This means that, no matter how large or small \(x\) gets, the value of \(y\) gets closer to zero without ever touching it.
Sketching Graphs
Sketching graphs combines the information of intercepts, symmetry, and asymptotic behavior to visualize a function's behavior. With all these insights, you can better predict the graph's shape.
Start by marking the intercepts, in this case at \((0,0)\). Knowing the graph's origin symmetry, you can expect it to mirror along the origin.
Start by marking the intercepts, in this case at \((0,0)\). Knowing the graph's origin symmetry, you can expect it to mirror along the origin.
- The asymptotic behavior showing \(y \rightarrow 0\) as \(|x|\) grows indicates the graph will flatten out near the x-axis.
- The symmetry about the origin tells us the graph appears in the opposite quadrant of any plotted part.
Other exercises in this chapter
Problem 22
$$ \text { In Problems 17-24, solve for } x . \text { Hint: } \log _{a} b=c \Leftrightarrow a^{c}=b \text {. } $$ $$ \log _{4}\left(\frac{1}{2 x}\right)=3 $$
View solution Problem 22
State whether each of the following is an odd function, an even function, or neither. Prove your statements. (a) The sum of two even functions (b) The sum of tw
View solution Problem 22
In Problems 15-30, specify whether the given function is even, odd, or neither, and then sketch its graph. \(\phi(z)=\frac{2 z+1}{z-1}\)
View solution Problem 22
Express the solution set of the given inequality in interval notation and sketch its graph. $$ (2 x+3)(3 x-1)(x-2)
View solution