Problem 12
Question
For the following exercises, find the domain, vertical asymptotes, and horizontes of the functions. $$ f(x)=\frac{x}{x^{2}-9} $$
Step-by-Step Solution
Verified Answer
Domain: all x except 3 and -3; vertical asymptotes at x = 3 and x = -3; horizontal asymptote at y = 0.
1Step 1: Identify Domain Restrictions
To find the domain of a function, determine what values of the variable x make the function undefined. For rational functions like \(f(x) = \frac{x}{x^2 - 9}\), check where the denominator equals zero. Set \(x^2 - 9 = 0\) and solve for \(x\).
2Step 2: Solve for x in Domain Restrictions
The equation \(x^2 - 9 = 0\) can be factored as \((x - 3)(x + 3) = 0\). Thus, \(x = 3\) or \(x = -3\). These are the values that make the function undefined, hence the domain is all real numbers except \(x = 3\) and \(x = -3\).
3Step 3: Find Vertical Asymptotes
Vertical asymptotes occur where the denominator is zero and the numerator is not zero simultaneously. From the solved domain restriction, the vertical asymptotes are at \(x = 3\) and \(x = -3\).
4Step 4: Determine Horizontal Asymptotes
For any rational function \(\frac{P(x)}{Q(x)}\), if the degree of \(P(x)\) is less than the degree of \(Q(x)\), the horizontal asymptote is \(y = 0\). Since the degree of \(x\) (numerator) is 1 and the degree of \(x^2 - 9\) (denominator) is 2, the horizontal asymptote is \(y = 0\).
Key Concepts
DomainVertical AsymptotesHorizontal Asymptotes
Domain
When exploring the domain of a rational function like \(f(x) = \frac{x}{x^2 - 9}\), the key is to identify the values of \(x\) that make the function undefined. This happens when the denominator equals zero. Thus, we set up the equation \(x^2 - 9 = 0\). By solving this equation, we determine the critical values where the function cannot exist.
To solve \(x^2 - 9 = 0\), notice that it factors neatly into \((x - 3)(x + 3) = 0\). Each factor can be set to zero, giving us \(x = 3\) and \(x = -3\). For the domain, we must exclude these values from all real numbers. Therefore, the domain of the function is all real numbers except \(x = 3\) and \(x = -3\). In interval notation, this is represented as \((-\infty, -3) \cup (-3, 3) \cup (3, \infty)\).
This process ensures that we only include the input values for which the function is well-defined, avoiding divisions by zero.
To solve \(x^2 - 9 = 0\), notice that it factors neatly into \((x - 3)(x + 3) = 0\). Each factor can be set to zero, giving us \(x = 3\) and \(x = -3\). For the domain, we must exclude these values from all real numbers. Therefore, the domain of the function is all real numbers except \(x = 3\) and \(x = -3\). In interval notation, this is represented as \((-\infty, -3) \cup (-3, 3) \cup (3, \infty)\).
This process ensures that we only include the input values for which the function is well-defined, avoiding divisions by zero.
Vertical Asymptotes
Vertical asymptotes in rational functions indicate where the function grows very large (positively or negatively). Specifically, they occur at the values of \(x\) that make the denominator zero while the numerator is non-zero, leading to undefined points where the graph shoots up to infinity.
From our earlier domain work, we found the critical values \(x = 3\) and \(x = -3\). These are where the denominator \(x^2 - 9\) becomes zero. Since \(x\) in the numerator is not zero at these points, vertical asymptotes are indeed present here. So, the vertical asymptotes are located at \(x = 3\) and \(x = -3\).
The presence of vertical asymptotes marks a vertical line on the graph where, as \(x\) approaches these values, \(f(x)\) correlates to extreme increases or decreases, achieving a near-vertical trajectory.
From our earlier domain work, we found the critical values \(x = 3\) and \(x = -3\). These are where the denominator \(x^2 - 9\) becomes zero. Since \(x\) in the numerator is not zero at these points, vertical asymptotes are indeed present here. So, the vertical asymptotes are located at \(x = 3\) and \(x = -3\).
The presence of vertical asymptotes marks a vertical line on the graph where, as \(x\) approaches these values, \(f(x)\) correlates to extreme increases or decreases, achieving a near-vertical trajectory.
Horizontal Asymptotes
Horizontal asymptotes provide insight into the end behavior of a rational function, detailing how the function behaves as \(x\) moves towards infinity or negative infinity.
For a rational function expressed as \(\frac{P(x)}{Q(x)}\), where \(P(x)\) and \(Q(x)\) are polynomials, the degrees of these polynomials direct us to the horizontal asymptote's location. If the degree of the numerator \(P(x)\) is less than the degree of the denominator \(Q(x)\), then the horizontal asymptote is \(y = 0\).
In our function \(f(x) = \frac{x}{x^2 - 9}\), the degree of the numerator \(x\) is 1, while the degree of the denominator \(x^2 - 9\) is 2. Since the numerator's degree is indeed less, we establish that the horizontal asymptote is \(y = 0\).
This horizontal line indicates that, as \(x\) travels to the bounds of infinity, the value of \(f(x)\) approaches zero, leveling off and sketching a flat path along the \(x\)-axis.
For a rational function expressed as \(\frac{P(x)}{Q(x)}\), where \(P(x)\) and \(Q(x)\) are polynomials, the degrees of these polynomials direct us to the horizontal asymptote's location. If the degree of the numerator \(P(x)\) is less than the degree of the denominator \(Q(x)\), then the horizontal asymptote is \(y = 0\).
In our function \(f(x) = \frac{x}{x^2 - 9}\), the degree of the numerator \(x\) is 1, while the degree of the denominator \(x^2 - 9\) is 2. Since the numerator's degree is indeed less, we establish that the horizontal asymptote is \(y = 0\).
This horizontal line indicates that, as \(x\) travels to the bounds of infinity, the value of \(f(x)\) approaches zero, leveling off and sketching a flat path along the \(x\)-axis.
Other exercises in this chapter
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