Verified Answer :
- Identify that the x-intercepts \((-4, 0)\) and \((4, 0)\) imply a numerator of \(x^2 - 16\).
- Recognize that the vertical asymptote at \(x = 2\) suggests a denominator of \(x - 2\).
- Use the oblique asymptote \(y = x + 2\) to confirm the form \(f(x) = (x + 2) + \frac{R}{x - 2}\), leading to \(f(x) = \frac{x^2 - 16}{x - 2}\).
- Conclude that the rational function is \(\boxed{f(x) = \frac{x^2 - 16}{x - 2}}\).
Explanation
- Analyzing the Given Information
We are looking for a rational function \(f(x)\) defined by its key features: x-intercepts at \((-4, 0)\) and \((4, 0)\), a vertical asymptote at \(x = 2\), and an oblique (slant) asymptote at \(y = x + 2\).
A rational function can be represented as the ratio of two polynomials, \(f(x) = \frac{N(x)}{D(x)}\).
The x-intercepts occur where the numerator \(N(x) = 0\), so the function must have factors \((x + 4)\) and \((x - 4)\) in the numerator.
The vertical asymptote occurs where the denominator is zero and the numerator is non-zero, so the denominator must contain a factor \((x - 2)\).
- Formulating the Function Expression
For a rational function to have an oblique asymptote, the degree of the numerator must be exactly one higher than the degree of the denominator. Since we have a vertical asymptote at \(x = 2\), the simplest denominator is \(D(x) = (x - 2)\).
Given the oblique asymptote \(y = x + 2\), we can express the function as:
\[f(x) = (x + 2) + \frac{R}{x - 2}\]
where \(R\) is a constant. We can rewrite this expression as a single rational function:
\[f(x) = \frac{(x + 2)(x - 2) + R}{x - 2} = \frac{x^2 - 4 + R}{x - 2}\]
- Determining and Verifying the Function
We know the numerator \(N(x) = x^2 - 4 + R\) must have roots at \(x = 4\) and \(x = -4\) because the x-intercepts are at \((-4, 0)\) and \((4, 0)\).
Setting \(N(4) = 0\):
\[4^2 - 4 + R = 0\]
\[16 - 4 + R = 0\]
\[12 + R = 0 \implies R = -12\]
Substituting \(R = -12\) back into our expression for \(f(x)\), we get:
\[f(x) = \frac{x^2 - 4 - 12}{x - 2} = \frac{x^2 - 16}{x - 2}\]
We can verify the intercepts: \(x^2 - 16 = 0\) when \(x = \pm 4\), confirming the x-intercepts. The vertical asymptote is at \(x = 2\), where the denominator is zero. Finally, performing polynomial division \(\frac{x^2 - 16}{x - 2}\):
\[x^2 - 16 = (x + 2)(x - 2) - 12\]
\[\frac{x^2 - 16}{x - 2} = x + 2 - \frac{12}{x - 2}\]
As \(x \to \infty\), the term \(-\frac{12}{x - 2} \to 0\), so the function approaches the line \(y = x + 2\), which is our oblique asymptote.
- Final Conclusion
The function is \(f(x) = \frac{x^2 - 16}{x - 2}\). The graph features two branches separated by the vertical asymptote \(x = 2\), x-intercepts at \((-4, 0)\) and \((4, 0)\), and follows the slant line \(y = x + 2\) for large values of \(x\). The final answer is \(\boxed{f(x) = \frac{x^2 - 16}{x - 2}}\).
Examples
Rational functions with oblique asymptotes are used in various fields, such as economics and engineering. For example, in a factory setting, the average cost per unit produced often follows a rational function. If a company has fixed costs and variable costs, the total cost function might lead to an average cost function with an oblique asymptote, representing the long-term cost trend as production volume increases. By identifying these features, managers can better predict long-term operational costs and make informed decisions about scaling production.