Problem 45
Question
Find an equation of a rational function \(f\) that satisfles the given conditions. vertical asymptote: \(x=4\) horizontal asymptote: \(y=-1\) \(x\) -intercept: 3
Step-by-Step Solution
Verified Answer
The rational function is \(f(x) = \frac{-x + 3}{x - 4}\).
1Step 1: Understand Vertical Asymptote Condition
A vertical asymptote at \(x = 4\) suggests the function has a factor of \((x - 4)\) in the denominator, because dividing by zero at \(x = 4\) causes the graph to approach infinity.
2Step 2: Analyze Horizontal Asymptote Condition
A horizontal asymptote at \(y = -1\) implies that as \(x\) approaches infinity, the values of \(f(x)\) approach \(-1\). For a rational function \(\frac{ax + b}{cx + d}\), if the degrees of numerator and denominator are the same, \(\lim_{{x \to \infty}} \frac{ax + b}{cx + d} = \frac{a}{c}\). Here, \(\frac{a}{c} = -1\) which gives us \(a = -c\).
3Step 3: Find the X-intercept Condition
An \(x\)-intercept of 3 means that when \(x = 3\), \(f(x) = 0\). This means the numerator must be zero, thus it has a factor of \((x - 3)\).
4Step 4: Construct the Rational Function
With the above information, we can write the rational function as \(f(x) = \frac{k(x - 3)}{(x - 4)}\). To satisfy the horizontal asymptote condition \(y = -1\), ensure that the leading coefficient of the numerator is the negation of the leading coefficient of the denominator, i.e., \(k(x - 3)\) should match \(-(x - 4)\). Therefore, \(f(x) = \frac{-(x - 3)}{x - 4}\) or simplified, \(f(x) = \frac{-x + 3}{x - 4}\).
Key Concepts
Vertical AsymptoteHorizontal AsymptoteX-Intercept
Vertical Asymptote
A vertical asymptote occurs in a rational function when the value of the function heads towards positive or negative infinity as the input approaches a certain value. In simpler terms, it's where the graph of the function shoots up or down without bounds.
For our function, the vertical asymptote at \(x = 4\) implies that the denominator of the rational function is zero when \(x = 4\). This causes the function to become undefined at \(x = 4\) and the graph to approach infinity.
To incorporate this into a function, include \((x - 4)\) in the denominator, ensuring that at \(x = 4\), division by zero occurs—a hallmark of a vertical asymptote.
For our function, the vertical asymptote at \(x = 4\) implies that the denominator of the rational function is zero when \(x = 4\). This causes the function to become undefined at \(x = 4\) and the graph to approach infinity.
To incorporate this into a function, include \((x - 4)\) in the denominator, ensuring that at \(x = 4\), division by zero occurs—a hallmark of a vertical asymptote.
- Vertical asymptote: \(x = 4\)
- Denominator factor: \((x - 4)\)
Horizontal Asymptote
In rational functions, a horizontal asymptote indicates the behavior of the function as \(x\) becomes very large or very small. It shows the value that \(f(x)\) approaches as \(x\) goes to infinity or negative infinity. For our specific case, the horizontal asymptote is \(y = -1\). This tells us how to balance the degrees and coefficients in our rational function.
The horizontal asymptote \(y = -1\) suggests that if our function is \(\frac{ax + b}{cx + d}\), then the leading coefficients should satisfy \(\lim_{{x \to \infty}} \frac{ax + b}{cx + d} = \frac{a}{c} = -1\). Here, we make sure that \(a\) and \(c\) are such that \(a = -c\).
In summary:
The horizontal asymptote \(y = -1\) suggests that if our function is \(\frac{ax + b}{cx + d}\), then the leading coefficients should satisfy \(\lim_{{x \to \infty}} \frac{ax + b}{cx + d} = \frac{a}{c} = -1\). Here, we make sure that \(a\) and \(c\) are such that \(a = -c\).
In summary:
- Horizontal asymptote: \(y = -1\)
- Condition: \(\frac{a}{c} = -1\)
- Solution: \(a = -c\)
X-Intercept
The \(x\)-intercept of a function is the value of \(x\) at which the function crosses the x-axis. This means the output of the function (\(f(x)\)) is zero at this point.
In our case, the \(x\)-intercept is 3. This means that when \(x = 3\), the function reaches zero. Thus, for the numerator of our rational function to be zero at this point, it must include \((x - 3)\) as a factor.
This ensures that:
In our case, the \(x\)-intercept is 3. This means that when \(x = 3\), the function reaches zero. Thus, for the numerator of our rational function to be zero at this point, it must include \((x - 3)\) as a factor.
This ensures that:
- When \(x = 3\), \(f(x) = 0\)
- Numerator factor: \((x - 3)\)
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