- Understand that a vertical asymptote in a rational function occurs at a specific \(x\)-value where the function is undefined.
- Recognize that vertical shifts of the form \(f(x) + k\) do not change the \(x\)-coordinates of a function's asymptotes.
- Identify that horizontal shifts change the location of vertical asymptotes.
- Conclude that the equation \(m(x) = u(x) + 7\) is the only transformation that keeps the discontinuity at \(x = 7\), as it only shifts the graph vertically.
The correct choice is \(\boxed{m(x) = u(x) + 7}\).
Explanation
- Analysis of the Problem
We are given that a rational function \(v(x)\) has a point of discontinuity at \(x = 7\). A point of discontinuity in a rational function typically occurs where the denominator equals zero, leading to a vertical asymptote at that \(x\)-value. We are looking for an equation for a function \(w(x)\) that also has a point of discontinuity at \(x = 7\). In the provided options, the notation uses \(u(x)\) as the known function (which we identify as \(v(x)\)) and \(m(x)\) as the new function \(w(x)\). We must test which transformation preserves the discontinuity at \(x = 7\).
- Evaluating Transformations of Discontinuities
Let \(v(x)\) be a rational function that is undefined at \(x = 7\). Transformations of a function \(f(x)\) affect its domain and range differently:
- A horizontal shift: \(f(x-h)\) shifts the graph to the right by \(h\), so the new discontinuity occurs at \(x = 7 + h\).
- A horizontal shift: \(f(x+h)\) shifts the graph to the left by \(h\), so the new discontinuity occurs at \(x = 7 - h\).
- A vertical shift: \(f(x) + k\) shifts the graph up by \(k\), which changes the \(y\)-values but keeps the \(x\)-value of the discontinuity at the same location.
- Combining shifts: \(f(x-h) + k\) moves the discontinuity to \(x = 7 + h\).
- Selecting the Correct Equation
Now, we evaluate the given options to see which preserves the discontinuity at \(x = 7\):
- \(m(x) = u(x-7)\): This shifts the graph right by 7, so the discontinuity occurs at \(x = 7 + 7 = 14\).
- \(m(x) = u(x+7)\): This shifts the graph left by 7, so the discontinuity occurs at \(x = 7 - 7 = 0\).
- \(m(x) = v(x-7) + 7\): This shifts the graph right by 7 and up by 7, so the discontinuity occurs at \(x = 7 + 7 = 14\).
- \(m(x) = u(x) + 7\): This shifts the graph only vertically by 7 units. The domain of the function remains the same, meaning it is still undefined at \(x = 7\). This preserves the point of discontinuity at \(x = 7\).
- Conclusion
Comparing our analysis with the provided options, we conclude that the transformation \(m(x) = u(x) + 7\) maintains the vertical asymptote at \(x = 7\). Therefore, this is the equation that correctly represents a function \(w(x)\) (represented as \(m(x)\)) with a discontinuity at \(x = 7\), given \(v(x)\) (represented as \(u(x)\)) also has one. The final answer is the fourth option.
Examples
In engineering or physics, vertical asymptotes often represent physical limits or points where a system becomes unstable, such as the voltage across a circuit component or the position of a particle under specific force fields. If an engineer knows that a system's output is unstable at a certain input value (e.g., \(x = 7\)), applying a vertical shift—like adding a constant baseline of \(7\) to the signal—maintains that instability at the exact same critical threshold, allowing for easier observation or data processing without changing the fundamental limitation of the system.