Rational Functions And Discontinuities Finding W(x) With Discontinuity At X=7

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Let's dive into the fascinating world of rational functions and their points of discontinuity. In this article, we'll explore how transformations can affect these discontinuities, specifically focusing on the behavior of functions $v(x)$ and $w(x)$ around the point $x=7$. We'll analyze the given options to determine which equation could represent function $w(x)$, given that both $v(x)$ and $w(x)$ have a point of discontinuity at $x=7$.

Understanding Rational Functions and Discontinuities

Before we tackle the problem, it's crucial to grasp the fundamentals. A rational function is essentially a function that can be expressed as the ratio of two polynomials, such as $f(x) = \frac{P(x)}{Q(x)}$, where $P(x)$ and $Q(x)$ are polynomials. These functions exhibit unique behaviors, and one of the most interesting is the presence of discontinuities.

Discontinuities occur at points where the function is not continuous, meaning there's a break or a jump in the graph. In rational functions, discontinuities typically arise when the denominator, $Q(x)$, equals zero. This is because division by zero is undefined, leading to a point where the function doesn't exist. These discontinuities can manifest as either vertical asymptotes or holes in the graph.

  • Vertical Asymptotes: These occur when the denominator is zero, and the numerator is non-zero at the same point. The function approaches infinity (or negative infinity) as $x$ approaches this value, creating a vertical line that the graph never crosses.
  • Holes (Removable Discontinuities): These happen when both the numerator and denominator are zero at the same point. This creates a "hole" in the graph, as the function is undefined at that specific $x$ value, but it can be "removed" by simplifying the rational expression.

Now, let's consider the given information. We know that both rational functions $v(x)$ and $w(x)$ have a point of discontinuity at $x=7$. This means that when $x=7$, the denominator of both functions becomes zero (or both numerator and denominator are zero, creating a hole). This is a crucial piece of information that will guide us in selecting the correct equation for $w(x)$.

Analyzing the Options: Horizontal Transformations

The core of this problem lies in understanding how transformations affect the graph of a function, particularly horizontal shifts. The given options involve transformations of the form $v(x ext{ ± } a)$, where $a$ is a constant. These transformations shift the graph of $v(x)$ horizontally. Let's break down the two main types of horizontal transformations:

  • Horizontal Shift to the Right: If we have a function $v(x - a)$, where $a > 0$, the graph of $v(x)$ is shifted a units to the right. This means that every point on the original graph is moved a units to the right.
  • Horizontal Shift to the Left: Conversely, if we have a function $v(x + a)$, where $a > 0$, the graph of $v(x)$ is shifted a units to the left. Every point on the original graph is moved a units to the left.

These transformations are pivotal in understanding how the discontinuity at $x=7$ might be affected. Remember, the discontinuity occurs because the denominator (or both numerator and denominator) of the function becomes zero at that point. Shifting the graph horizontally will shift the location of this zero, and thus, the location of the discontinuity.

Let's analyze option A: $w(x) = v(x - 7)$. This represents a horizontal shift of the graph of $v(x)$ 7 units to the right. If $v(x)$ has a discontinuity at $x=7$, this means the denominator of $v(x)$ (or both numerator and denominator) is zero when $x=7$. Now, consider $w(x) = v(x - 7)$. For $w(x)$ to have a discontinuity, the expression inside the function, $x - 7$, must be equal to the value that causes a discontinuity in $v(x)$. Since $v(x)$ has a discontinuity at $x=7$, we need $x - 7 = 7$ to create a similar discontinuity in $w(x)$. This means $x = 14$. Therefore, $w(x)$ will have a discontinuity at $x=14$, not at $x=7$. This eliminates option A.

Now, let's examine option B: $w(x) = v(x + 7)$. This represents a horizontal shift of the graph of $v(x)$ 7 units to the left. Again, $v(x)$ has a discontinuity at $x=7$, meaning its denominator (or both numerator and denominator) is zero when $x=7$. For $w(x) = v(x + 7)$ to have a discontinuity, the expression $x + 7$ must be equal to 7. So, we need $x + 7 = 7$, which implies $x = 0$. Therefore, $w(x)$ will have a discontinuity at $x=0$, not at $x=7$. This also eliminates option B.

The Key Insight: A Deeper Look at the Discontinuity

So far, we've analyzed horizontal shifts, but we haven't found an option that maintains the discontinuity at $x=7$. This suggests that the correct equation might involve a different kind of transformation, or perhaps a more direct relationship between $v(x)$ and $w(x)$. To understand this, we need to consider what it truly means for a rational function to have a discontinuity at $x=7$.

As we discussed earlier, a discontinuity arises when the denominator of the rational function is zero. Let's say $v(x)$ has a denominator of the form $(x - 7) imes ext{some other factor}$. This ensures that $v(x)$ has a discontinuity at $x=7$. Now, for $w(x)$ to also have a discontinuity at $x=7$, its denominator must also have a factor of $(x - 7)$.

Consider what would happen if we were to simply add a constant to $v(x)$, or multiply it by a constant. These transformations (vertical shifts and vertical stretches/compressions) would not change the location of the zeros in the denominator. Therefore, if $v(x)$ has a discontinuity at $x=7$, these transformations would not shift the location of that discontinuity. However, these options are not provided in the questions.

Let's think about what might happen if $v(x)$ has a removable discontinuity (a hole) at $x=7$. This means that both the numerator and denominator of $v(x)$ have a factor of $(x-7)$. If we construct $w(x)$ such that it also has the same factor in the numerator and the denominator, it would also have a hole at $x=7$.

Now, let's reconsider the options. Options A and B involve horizontal shifts, which we've already ruled out. Let's construct a potential scenario to check our hypothesis. Suppose $v(x) = \frac{x-7}{(x-7)(x+2)}$. This function has a removable discontinuity at $x=7$ and a vertical asymptote at $x=-2$.

If $w(x)$ also has a discontinuity at $x=7$, it could potentially be represented by a function with a similar form, perhaps with a different numerator or additional factors in the denominator, as long as the factor $(x-7)$ remains.

Refining the Problem Question

The original problem asks "Which equation could represent function $w(x)$?". This wording is crucial. It doesn't ask for the only equation that could represent $w(x)$, but rather one possibility. This opens the door to multiple potential solutions, as long as the condition of a discontinuity at $x=7$ is met.

Considering our analysis, options A and B shift the discontinuity away from $x=7$, making them incorrect. To solidify our understanding, let's think of a more direct transformation that would maintain the discontinuity at $x=7$. For instance, if $w(x)$ is equal to $v(x)$ multiplied by a constant or added to a function that does not have a discontinuity at $x=7$, then the discontinuity at $x=7$ would be retained.

Conclusion: Identifying the Correct Approach

In conclusion, the key to solving this problem is understanding how transformations affect the discontinuities of rational functions. Horizontal shifts, as represented in options A and B, will move the location of the discontinuity. Therefore, the correct answer must involve a transformation or relationship that preserves the discontinuity at $x=7$. The problem as it is listed does not provide an answer that preserves the discontinuity at x=7. Therefore, more options must be provided to get a valid answer.

Original Question: Rational functions $v$ and $w$ both have a point of discontinuity at $x=7$. Which equation could represent function $w$?

Rewritten Question: Suppose rational functions $v(x)$ and $w(x)$ both exhibit a point of discontinuity at $x=7$. Which of the following equations could potentially represent the function $w(x)$, considering the relationship with $v(x)$?

Rational Functions and Discontinuities: Finding w(x) with Discontinuity at x=7