- The denominator of the function \(f(x) = \frac{x-5}{3x^2 - 17x - 28}\) is factored as \((3x+4)(x-7)\).
- Discontinuities are identified at \(x=7\) and \(x=-\frac{4}{3}\) where the denominator is zero.
- Since the numerator \((x-5)\) shares no common factors with the denominator, no holes exist; instead, vertical asymptotes occur at these points.
- The correct statement is: there are asymptotes at \(x=7\) and \(x=-\frac{4}{3}\).
- Final Answer: \(\boxed{\text{There are asymptotes at } x=7 \text{ and } x=-\frac{4}{3}}\)
Explanation
- Analyzing the Function and Finding Discontinuities
The function is given by \(f(x) = \frac{x-5}{3x^2 - 17x - 28}\). To understand the nature of its discontinuities, we first identify the values of \(x\) for which the function is undefined by setting the denominator equal to zero: \(3x^2 - 17x - 28 = 0\). We can factor this quadratic expression by looking for two numbers that multiply to \(3 \times (-28) = -84\) and add to \(-17\). These numbers are \(-21\) and \(4\). Therefore, we rewrite the middle term and factor by grouping:
\[3x^2 - 21x + 4x - 28 = 3x(x-7) + 4(x-7) = (3x+4)(x-7)\]
Setting each factor equal to zero, we find the roots of the denominator to be \(x = 7\) and \(x = -\frac{4}{3}\). These are the points where the function has potential discontinuities.
- Distinguishing Between Holes and Asymptotes
A discontinuity at a point \(x = c\) is considered a "hole" (or removable discontinuity) if the factor \((x-c)\) is present in both the numerator and the denominator, allowing it to be canceled out. If the factor remains in the denominator after any possible simplification, the discontinuity is an infinite discontinuity, represented by a vertical asymptote at \(x = c\). Looking at our function \(f(x) = \frac{x-5}{(3x+4)(x-7)}\), the numerator is \((x-5)\). Since \((x-5)\) is not a factor of the denominator, there are no common factors to cancel. Therefore, the function cannot be simplified to remove the division by zero at either point.
- Final Conclusion Regarding Discontinuities
Because the factors \((x-7)\) and \((3x+4)\) in the denominator do not cancel with anything in the numerator, the function approaches infinity as \(x\) approaches \(7\) or \(-\frac{4}{3}\). Consequently, both points result in vertical asymptotes. We can compare this with the given choices, and the correct statement is that there are vertical asymptotes at \(x=7\) and \(x=-\frac{4}{3}\).
Examples
In electrical engineering, engineers must identify the locations of poles (where the denominator is zero) in a transfer function to analyze the stability of a circuit. If a pole corresponds to a value of a parameter like frequency, it indicates a resonant frequency where the system's output could theoretically approach infinity, which corresponds to the concept of a vertical asymptote. Identifying whether these points are removable holes or essential vertical asymptotes helps engineers design stable systems that avoid these destructive resonances.