Chapter 2: Problem 106
Graph the curves. Explain the relationship between the curve's formula and what you see. $$y=\frac{-1}{\sqrt{4-x^{2}}}$$
Short Answer
Expert verified
The graph is an inverted U-shape between -2 and 2, symmetric about the y-axis, approaching negative infinity at the ends.
Step by step solution
01
Identify the domain of the function
The function given is \( y = \frac{-1}{\sqrt{4-x^2}} \). We need to ensure that the denominator is not zero and that we are not taking the square root of a negative number. Hence, we require \( 4 - x^2 > 0 \). Solving this inequality gives \( -2 < x < 2 \). Thus, the domain of the function is the open interval \( (-2, 2) \).
02
Analyze the vertical asymptotes
Since the function's denominator \( \sqrt{4-x^2} \) approaches zero when \( x \) approaches \( -2 \) or \( 2 \), there are vertical asymptotes at \( x = -2 \) and \( x = 2 \). The function will not exist beyond these points due to division by zero.
03
Determine the behavior near vertical asymptotes
As \( x \) approaches \( -2 \) from the right, \( \sqrt{4-x^2} \) approaches zero, which means \( y = \frac{-1}{\sqrt{4-x^2}} \) approaches negative infinity. Likewise, as \( x \) approaches \( 2 \) from the left, \( y \) also approaches negative infinity.
04
Find symmetry
The function is composed of even powers of \( x \), so we can test for even symmetry by seeing if \( f(-x) = f(x) \). Compute \( y = -\frac{1}{\sqrt{4 - (-x)^2}} = -\frac{1}{\sqrt{4-x^2}} \), which is the same as the original function. Therefore, the function is symmetric with respect to the y-axis.
05
Sketch the graph
From the analysis, draw vertical asymptotes at \( x = -2 \) and \( x = 2 \). Within the interval \((-2, 2)\), the function is below the x-axis and symmetric around the y-axis, approaching negative infinity as it nears the asymptotes. The curve looks roughly like an 'inverted' U-shape within the domain.
06
Explain the visual representation
The curve represents a hyperbolic function that is symmetrical about the y-axis, with vertical asymptotes at \( x = -2 \) and \( x = 2 \). It mirrors across the y-axis and approaches negative infinity as it nears the asymptotes, consistent with its formula.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Domain of a Function
When we talk about the domain of a function, we're referring to all the possible input values (or "x" values) that will give us a real output. For the function \( y = \frac{-1}{\sqrt{4-x^2}} \), we have to ensure two things:
1. The denominator \( \sqrt{4-x^2} \) must not be zero, as division by zero is undefined.
2. The expression under the square root \( 4-x^2 \) must be positive, because taking the square root of a negative number isn't defined in the real number system.
To find where \( \sqrt{4-x^2} \) is positive, we solve the inequality \( 4-x^2 > 0 \).
- Rewriting it, we get \( -2 < x < 2 \).
This means the domain, or the set of all possible \( x \)-values input into the function that yield a real result, is the open interval \( (-2, 2) \). "Open interval" means \( x \) cannot be \(-2\) or \(2\) exactly, just all the values between these points.
1. The denominator \( \sqrt{4-x^2} \) must not be zero, as division by zero is undefined.
2. The expression under the square root \( 4-x^2 \) must be positive, because taking the square root of a negative number isn't defined in the real number system.
To find where \( \sqrt{4-x^2} \) is positive, we solve the inequality \( 4-x^2 > 0 \).
- Rewriting it, we get \( -2 < x < 2 \).
This means the domain, or the set of all possible \( x \)-values input into the function that yield a real result, is the open interval \( (-2, 2) \). "Open interval" means \( x \) cannot be \(-2\) or \(2\) exactly, just all the values between these points.
Vertical Asymptotes
Vertical asymptotes are lines where a function approaches infinity, and they occur where the denominator of a rational function becomes zero but the numerator doesn't. For our specific function \( y = \frac{-1}{\sqrt{4-x^2}} \), we see that the denominator \( \sqrt{4-x^2} \) becomes zero when \( x = -2 \) and \( x = 2 \).
- At these \( x \) values, the function "blows up," meaning it heads towards negative infinity since the negative sign in the numerator does not change.
- Thus, we have vertical asymptotes at \( x = -2 \) and \( x = 2 \).
Symmetry in Functions
Understanding symmetry in functions can simplify graphing and analyzing them. A function is symmetric about the y-axis if it remains unchanged when \( x \) is replaced by \(-x\). This type of symmetry is called "even symmetry."
For the function \( y = \frac{-1}{\sqrt{4-x^2}} \), by calculating \( f(-x) = -\frac{1}{\sqrt{4-(-x)^2}} \), we find it equals \( -\frac{1}{\sqrt{4-x^2}} \), which is identical to the original function. Hence, this function is even and symmetric about the y-axis.
For the function \( y = \frac{-1}{\sqrt{4-x^2}} \), by calculating \( f(-x) = -\frac{1}{\sqrt{4-(-x)^2}} \), we find it equals \( -\frac{1}{\sqrt{4-x^2}} \), which is identical to the original function. Hence, this function is even and symmetric about the y-axis.
- This symmetry implies that the left half of the graph is a mirror image of the right half across the y-axis.
- It further means that understanding one side of the function often gives you insight into the entire shape.