- Identify the denominator of the function \(f(x) = \frac{14}{(x-5)(x+1)}\).
- Set the denominator equal to zero: \((x-5)(x+1) = 0\).
- Solve for \(x\) to find the vertical asymptotes, which are \(x = 5\) and \(x = -1\).
- Select the matching option: \(\boxed{x=-1 \text{ and } x=5}\).
Explanation
- Analysis of the Rational Function
To find the vertical asymptotes of a rational function \(f(x) = \frac{N(x)}{D(x)}\), we identify the values of \(x\) where the denominator \(D(x)\) equals zero, provided the numerator \(N(x)\) is not zero at those same values. In this problem, we have the function \(f(x) = \frac{14}{(x-5)(x+1)}\). Here, the numerator is the constant \(N(x) = 14\) and the denominator is \(D(x) = (x-5)(x+1)\).
- Finding Potential Asymptotes
We set the denominator to zero to find the points where the function might have vertical asymptotes: \((x-5)(x+1) = 0\). This equation is satisfied when either factor is zero. Setting the first factor to zero, we get \(x - 5 = 0\), which results in \(x = 5\). Setting the second factor to zero, we get \(x + 1 = 0\), which results in \(x = -1\).
- Verification and Conclusion
For both \(x = 5\) and \(x = -1\), we must check the numerator \(N(x) = 14\). Since the numerator is a constant value of 14, it is never equal to zero. Therefore, the function \(f(x)\) has vertical asymptotes at both \(x = 5\) and \(x = -1\). Comparing these results with the given options, we find that the first option, \(x = -1\) and \(x = 5\), is correct.
Examples
Vertical asymptotes are essential in fields like electrical engineering and physics when studying resonance or feedback control systems. For example, in a circuit, a function representing gain might have an asymptote at a specific frequency, signaling where the gain approaches infinity, which could indicate a point of system instability or resonance. Identifying these points mathematically helps engineers predict and avoid such critical conditions.