Verified Answer :
- Simplify the function \(f(x) = \frac{-(x + 6)(x - 3)}{(x + 3)(x - 3)}\) to \(f(x) = \frac{-x - 6}{x + 3}\) for \(x eq 3\).
- Identify that there is a hole at \(x = 3\) and a vertical asymptote at \(x = -3\).
- Calculate the y-intercept by evaluating \(f(0) = -2\) and the x-intercept by setting the numerator of the simplified function to zero, yielding \(x = -6\).
- Determine that the horizontal asymptote is \(y = -1\) because the limit as \(x \to \infty\) is \(-1\).
- The true statements are the y-intercept at \(-2\), the x-intercept at \(-6\), and the vertical asymptote at \(x = -3\). The final answer is \(\boxed{\text{y-intercept at } -2, \text{ x-intercept at } -6, \text{ vertical asymptote at } x = -3}\).
Explanation
- Factoring the Function
We begin by analyzing the function \(f(x) = \frac{18 - 3x - x^2}{x^2 - 9}\) to identify its key features. First, we factor both the numerator and the denominator:
The numerator is \(18 - 3x - x^2 = -(x^2 + 3x - 18) = -(x + 6)(x - 3)\).
The denominator is \(x^2 - 9 = (x - 3)(x + 3)\).
Thus, the function can be written as \(f(x) = \frac{-(x + 6)(x - 3)}{(x + 3)(x - 3)}\).
- Simplifying the Function and Identifying Holes
We observe a common factor of \((x - 3)\) in both the numerator and the denominator. For all \(x
eq 3\), we can cancel this factor:
\(f(x) = \frac{-(x + 6)}{x + 3} = \frac{-x - 6}{x + 3}\).
Because the factor \((x - 3)\) was canceled, the function has a removable discontinuity (a hole) at \(x = 3\). Therefore, the graph does not have a vertical asymptote at \(x = 3\).
- Finding Intercepts
To find the y-intercept, we evaluate the function at \(x = 0\):
\(f(0) = \frac{-(0 + 6)}{0 + 3} = \frac{-6}{3} = -2\).
Thus, the y-intercept is indeed at \(-2\).
To find the x-intercepts, we set the numerator of the simplified function to zero and solve for \(x\):
\(-(x + 6) = 0 \implies x = -6\).
Since \(x = -6\) is not equal to the values where the function is undefined (\(x = 3\) or \(x = -3\)), the graph has an x-intercept at \(-6\).
- Determining Asymptotes
Vertical asymptotes occur where the denominator is zero after simplification. Setting the simplified denominator to zero gives \(x + 3 = 0\), which results in \(x = -3\). So, there is a vertical asymptote at \(x = -3\).
For the horizontal asymptote, we consider the behavior of the function as \(x\) approaches infinity:
\(\lim_{x \to \infty} f(x) = \lim_{x \to \infty} \frac{-x - 6}{x + 3} = \lim_{x \to \infty} \frac{-1 - 6/x}{1 + 3/x} = -1\).
Thus, the horizontal asymptote is at \(y = -1\), not \(y = 1\).
- Final Conclusion
Based on our analysis, the true statements are:
- The graph has a y-intercept at \(-2\).
- The graph has an x-intercept at \(-6\).
- The graph has a vertical asymptote at \(x = -3\).
The other statements regarding a vertical asymptote at \(x = 3\) and a horizontal asymptote at \(y = 1\) are incorrect.
Examples
Understanding the properties of rational functions like finding asymptotes and intercepts is crucial in fields such as engineering and physics, where these functions model physical phenomena like sound intensity or electronic resistance. For example, a vertical asymptote might represent a condition where a system's output goes to infinity, signifying a potential failure point or a state of resonance that engineers must account for in their designs. Analyzing \(f(x) = \frac{18 - 3x - x^2}{x^2 - 9}\) teaches students how to pinpoint these critical operational limits in complex mathematical models.