Problem 30
Question
In Exercises \(27-34,\) (a) find the vertical asymptotes of the graph of \(f(x) .(\) b) Describe the behavior of \(f(x)\) to the left and right of each vertical asymptote. $$f(x)=\frac{1-x}{2 x^{2}-5 x-3}$$
Step-by-Step Solution
Verified Answer
The vertical asymptotes of the function \(f(x)=\frac{1-x}{2x^{2}-5x-3}\) are at \(x = 3\) and \(x = -0.5\). Close to these points, the function approaches \(-\infty\), from both left and right sides.
1Step 1: Find the Vertical Asymptotes
To find the vertical asymptotes of a function, we need to determine the values of \(x\) for which the function is undefined, i.e., when the denominator equals zero. So, we start by equating the denominator \(2x^{2}-5x-3\) to zero and solve for \(x\). This gives us the quadratic equation: \(2x^{2}-5x-3 = 0\). Solving this quadratic equation we get \(x = frac{- b \pm \sqrt {b^{2}-4ac}}{2a}\) where \(a = 2, b = -5, c = -3\). Therefore \(x = frac{5 \pm \sqrt {(-5)^{2}-4*2*(-3)}}{2*2}\) yields \(x = 3, -0.5\). So vertical asymptotes are at \(x = 3, -0.5\).
2Step 2: Describe the Behavior of f(x) to the Left and Right of Each Vertical Asymptote
To describe the behavior of \(f(x)\) near the vertical asymptotes, analyze the function values at points slightly to the left and right of the asymptotes. Using limits, for \(x = 3\), we find \(\lim_{x \to 3^-} f(x)\) and \(\lim_{x \to 3^+} f(x)\), similarly for \(x = -0.5\), we find \(\lim_{x \to -0.5^-} f(x)\) and \(\lim_{x \to -0.5^+} f(x)\). After calculating, we can see the function approaches \(+\infty\) or \(-\infty\) at different sides of the asymptote. Near \(x = 3\), the function approaches \(-\infty\) from the left and \(-\infty\) from the right. Near \(x = -0.5\), the function approaches \(-\infty\) from the left and \(-\infty\) from the right.
Key Concepts
Understanding Quadratic EquationsExploring Limits in Asymptotic AnalysisUnderstanding Asymptotic Behavior
Understanding Quadratic Equations
Quadratic equations are a type of polynomial equation that can be written in the form of \(ax^2 + bx + c = 0\), where \(a\), \(b\), and \(c\) are constants, and \(a eq 0\).
These equations are of utmost importance in finding vertical asymptotes because they help determine the values of \(x\) where the denominator of a rational function becomes zero, making the function undefined.
When solving a quadratic equation, one of the most common methods uses is the quadratic formula:
In the problem given, solving the quadratic equation \(2x^2 - 5x - 3 = 0\) yields two solutions: \(x = 3\) and \(x = -0.5\).
These values represent the points where vertical asymptotes occur.
These equations are of utmost importance in finding vertical asymptotes because they help determine the values of \(x\) where the denominator of a rational function becomes zero, making the function undefined.
When solving a quadratic equation, one of the most common methods uses is the quadratic formula:
- \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)
In the problem given, solving the quadratic equation \(2x^2 - 5x - 3 = 0\) yields two solutions: \(x = 3\) and \(x = -0.5\).
These values represent the points where vertical asymptotes occur.
Exploring Limits in Asymptotic Analysis
Limits play a crucial role in analyzing the behavior of functions, especially near points where they become undefined, such as vertical asymptotes.
By evaluating limits, we can predict how a function behaves as it approaches a certain point.
For vertical asymptotes, the limits on either side (left and right) are evaluated to understand whether the function approaches positive or negative infinity, indicating the asymptotic behavior.To illustrate using the provided function \(f(x) = \frac{1-x}{2x^2 - 5x - 3}\):
By evaluating limits, we can predict how a function behaves as it approaches a certain point.
For vertical asymptotes, the limits on either side (left and right) are evaluated to understand whether the function approaches positive or negative infinity, indicating the asymptotic behavior.To illustrate using the provided function \(f(x) = \frac{1-x}{2x^2 - 5x - 3}\):
- For the asymptote at \(x = 3\), we compute \(\lim_{x \to 3^-} f(x)\) and \(\lim_{x \to 3^+} f(x)\).
- This reveals how the function diverges or converges as it nears \(x = 3\), resulting in specific limits indicating the direction towards positive or negative infinity.
Understanding Asymptotic Behavior
Asymptotic behavior describes how functions behave as they approach specific points in their domain—particularly points where they become undefined, such as vertical asymptotes.
When a function's output seems to head towards infinity (positive or negative), it signals the presence of a vertical asymptote.In our exercise, it's essential to note how \(f(x)\) acts near its vertical asymptotes:
When a function's output seems to head towards infinity (positive or negative), it signals the presence of a vertical asymptote.In our exercise, it's essential to note how \(f(x)\) acts near its vertical asymptotes:
- At \(x = 3\), the function approaches \(-\infty\) both from the left and the right.
- Similarly, at \(x = -0.5\), the behavior is \(-\infty\) from both directions.
Other exercises in this chapter
Problem 30
Horizontal Tangent At what point is the tangent to \(f(x)=3-4 x-x^{2}\) horizontal?
View solution Problem 30
In Exercises 29 and 30 , use a graph to show that the limit does not exist. $$\lim _ { x \rightarrow 2 } \frac { x + 1 } { x ^ { 2 } - 4 }$$
View solution Problem 31
Finding Tangents and Normals (a) Find an equation for each tangent to the curve \(y=1 /(x-1)\) that has slope \(-1 .\) (See Exercise \(21 )\) (b) Find an equati
View solution Problem 31
In Exercises 31 and \(32,\) explain why the given function is continuous. $$f(x)=\frac{1}{x-3}$$
View solution