Analyzing The Function $f(x) = \frac{2x-5}{2(x^2-1)}$

Let's dive into the function $f(x) = \frac{2x-5}{2(x^2-1)}$ and dissect its properties. Understanding the behavior of rational functions like this one is fundamental in mathematics, allowing us to predict where they might intersect axes, where they are undefined, and how they behave across different intervals. We'll systematically analyze the provided statements to determine which ones accurately describe this specific function. Our journey will involve looking at zeros, $x$-intercepts, and vertical asymptotes, which are key characteristics of any function.

Understanding Zeros and $x$-Intercepts

When we talk about the zeros of a function, we're referring to the values of $x$ for which $f(x) = 0$. For a rational function, $f(x) = \frac{P(x)}{Q(x)}$, the zeros occur when the numerator, $P(x)$, is equal to zero, provided that the denominator, $Q(x)$, is not also zero at those same values of $x$. In our case, the numerator is $2x - 5$. Setting this to zero, we get $2x - 5 = 0$, which leads to $2x = 5$, and therefore $x = \frac{5}{2}$. So, the only real zero of this function is $x = \frac{5}{2}$.

Now, let's consider the concept of $x$-intercepts. An $x$-intercept is a point where the graph of the function crosses or touches the $x$-axis. This happens when $y = f(x) = 0$. Therefore, the $x$-coordinates of the $x$-intercepts are precisely the real zeros of the function. Since the only real zero of $f(x)$ is $x = \frac{5}{2}$, the only $x$-intercept is the point $(\frac{5}{2}, 0)$.

Let's examine the given statements in light of this understanding:

• Statement A: $f(x)$ has real zeros at 1 and -1. Based on our analysis, the only real zero is at $x = \frac{5}{2}$. Therefore, this statement is false.
• Statement B: $f(x)$ crosses the $x$-axis when $x=1$ and $x=-1$. Crossing the $x$-axis occurs at the $x$-intercepts, which correspond to the real zeros. Since $x=1$ and $x=-1$ are not zeros of the function, this statement is also false.
• Statement C: $f(x)$ has $x$-intercepts at $(1,0)$ and $(-1,0)$. As we've determined, the only $x$-intercept is at $(\frac{5}{2}, 0)$. Thus, this statement is false.

It seems there might be a misunderstanding in the provided options, as none of them accurately describe the zeros or $x$-intercepts of the function $f(x) = \frac{2x-5}{2(x^2-1)}$. Let's proceed to analyze other key features of this function, such as its vertical asymptotes, to provide a more complete picture.

Investigating Vertical Asymptotes

Vertical asymptotes are vertical lines that the graph of a function approaches but never touches. For a rational function $f(x) = \frac{P(x)}{Q(x)}$, vertical asymptotes occur at the values of $x$ where the denominator $Q(x)$ is equal to zero, and the numerator $P(x)$ is not zero at those same values. If both the numerator and denominator are zero at a certain $x$-value, it indicates a hole in the graph or a removable discontinuity, rather than a vertical asymptote.

In our function, $f(x) = \frac{2x-5}{2(x^2-1)}$, the denominator is $2(x^2-1)$. We need to find the values of $x$ that make this denominator zero:

$2(x^2-1) = 0$

Dividing by 2, we get:

$x^2-1 = 0$

This is a difference of squares, which can be factored as:

$(x-1)(x+1) = 0$

This equation holds true when $x-1=0$ or $x+1=0$. Therefore, the solutions are $x = 1$ and $x = -1$.

Now, we must check if the numerator, $2x-5$, is zero at these values.

• At $x=1$: The numerator is $2(1) - 5 = 2 - 5 = -3$. Since $-3 eq 0$, there is a vertical asymptote at $x = 1$.
• At $x=-1$: The numerator is $2(-1) - 5 = -2 - 5 = -7$. Since $-7 eq 0$, there is a vertical asymptote at $x = -1$.

So, the function $f(x) = \frac{2x-5}{2(x^2-1)}$ has vertical asymptotes at $x=1$ and $x=-1$.

Re-evaluating the Statements

Given our thorough analysis of zeros, $x$-intercepts, and vertical asymptotes, let's revisit the original statements. It's possible the question intended to test understanding of asymptotes, or there's a typo in the options provided.

Let's assume the question implicitly expects us to identify characteristics related to where the function is undefined or behaves singularly. Based on our findings:

• The function is undefined at $x=1$ and $x=-1$ because these values make the denominator zero.
• These points ($x=1$ and $x=-1$) correspond to vertical asymptotes, not zeros or $x$-intercepts.

If we were to hypothesize about what the intended correct answers might have been, perhaps they related to the domain or points of discontinuity. The domain of $f(x)$ is all real numbers except $x=1$ and $x=-1$.

Let's consider alternative interpretations or common misconceptions related to rational functions:

1. Confusion between zeros and values where the denominator is zero: Students might mistakenly think that if a value makes the denominator zero, it's a zero of the function. This is incorrect.
2. Misinterpreting asymptotes as intercepts: It's crucial to differentiate between where a graph crosses the $x$-axis (zeros/intercepts) and where it shoots off towards infinity (asymptotes).

Let's assume, for the sake of providing a comprehensive answer, that the question might have been phrased differently or the options were meant to reflect the asymptotes. If the question had asked about vertical asymptotes, then statements describing these would be correct. For instance, a statement like "$f(x)$ has vertical asymptotes at $x=1$ and $x=-1$" would be accurate.

However, strictly adhering to the given options and the definition of zeros and $x$-intercepts, none of the statements A, B, or C are correct. Let's imagine the question was a multiple-choice question where you must choose three correct answers. This scenario strongly suggests that the provided options are not directly about the function f(x)=rac{2x-5}{2(x^2-1)} as stated, or there's a significant error in the question's formulation or the options themselves. It's a common situation in test preparation where errors occur.

If we are forced to select three statements and assume there's a hidden context or a mistake, let's analyze the structure of the statements:

• Statements A and B talk about "real zeros" and "crossing the x-axis" at specific points. These are fundamentally about the numerator being zero.
• Statement C talks about "x-intercepts" at specific points, which is also about the numerator being zero.

Since the function $f(x) = \frac{2x-5}{2(x^2-1)}$ has its numerator $2x-5$ equal to zero only at $x = 5/2$, none of these statements can be correct. The denominator $2(x^2-1)$ is zero at $x=1$ and $x=-1$. These are the locations of vertical asymptotes, not zeros or $x$-intercepts. The graph of the function will approach infinity as $x$ approaches $1$ or $-1$. It will never actually touch or cross the $x$-axis at $x=1$ or $x=-1$.

Conclusion

After a thorough mathematical examination, the function $f(x) = \frac{2x-5}{2(x^2-1)}$ has a single real zero at $x = \frac{5}{2}$, resulting in one $x$-intercept at $(\frac{5}{2}, 0)$. It possesses vertical asymptotes at $x=1$ and $x=-1$. Based on these facts, none of the provided statements (A, B, and C) accurately describe the function's zeros or $x$-intercepts. It is highly probable that there is an error in the question or the given options.

If this were a real test scenario and you were required to select three options, it would indicate a flawed question. However, in a learning context, it's important to understand why these statements are incorrect. They incorrectly identify the points where the function's denominator is zero ($x=1, x=-1$) as points where the function equals zero (zeros/x-intercepts). For accurate mathematical analysis, always remember to find the zeros of the numerator and exclude values that also make the denominator zero.

For further understanding of rational functions and their properties, you can explore resources like Khan Academy's section on rational functions, which offers detailed explanations and examples.