Graphing Rational Functions: F(x) = 2x / (x^2 - 1)

Hey math enthusiasts! Today, we're diving deep into the wild world of rational functions, specifically tackling the beast that is f(x)=rac{2 x}{x^2-1}. You know, those functions with polynomials in both the numerator and the denominator? Yeah, those guys can be tricky. But fear not! By breaking them down step-by-step, we'll figure out exactly which graph represents this function. It's all about understanding the key features: domain, vertical asymptotes, horizontal or slant asymptotes, x-intercepts, and y-intercepts. Mastering these will turn you into a graph-predicting wizard, I promise! So grab your pencils, your calculators, and let's get ready to conquer this function.

Understanding the Domain and Vertical Asymptotes

First things first, guys, we need to nail down the domain of our function, f(x)=rac{2 x}{x^2-1}. The domain is basically all the possible x-values that won't make our function go haywire. For rational functions, the main culprit is division by zero. So, we need to find where the denominator, $x^2-1$, equals zero. Setting $x^2-1 = 0$ gives us $x^2 = 1$, which means $x = 1$ and $x = -1$. These are the values that are not in our domain. Therefore, our domain is all real numbers except $x=1$ and $x=-1$. Now, these excluded x-values are super important because they usually signal the presence of vertical asymptotes. These are vertical lines on our graph that the function approaches but never actually touches. So, we've got two vertical asymptotes at $x = 1$ and $x = -1$. Keep these in mind; they're like the boundaries of our graphing adventure. When sketching the graph, you'll see the function's behavior getting wild as it gets closer and closer to these lines, heading towards positive or negative infinity.

Finding Horizontal or Slant Asymptotes

Next up, let's talk about asymptotes, specifically horizontal or slant ones. These tell us about the function's behavior as x heads off towards positive or negative infinity. For our function, f(x)=rac{2 x}{x^2-1}, we compare the degree of the numerator (which is 1, from the $2x$) to the degree of the denominator (which is 2, from the $x^2$). Here's the golden rule, guys: If the degree of the denominator is greater than the degree of the numerator, then the horizontal asymptote is always the x-axis, which is the line $y=0$. And that's exactly our situation! The degree of $x^2-1$ (which is 2) is greater than the degree of $2x$ (which is 1). So, our horizontal asymptote is $y=0$. This means as x gets super large in either the positive or negative direction, our function's graph will get closer and closer to the x-axis. It won't cross it, but it'll hug it. This information is crucial for understanding the overall shape and long-term trend of the graph. It helps us predict what the function is doing way out there on the edges of our graph paper.

Identifying X-Intercepts and Y-Intercepts

Now, let's find where our function actually touches the axes – the intercepts. To find the x-intercepts, we set the numerator equal to zero, because that's the only way a fraction can be zero (as long as the denominator isn't also zero at that point, which we've already checked!). So, we set $2x = 0$, which gives us $x = 0$. This means our graph crosses the x-axis at the point $(0, 0)$. Easy peasy, right? For the y-intercept, we plug in $x = 0$ into the function. f(0) = rac{2(0)}{(0)^2-1} = rac{0}{-1} = 0. So, our graph also crosses the y-axis at the point $(0, 0)$. It turns out our function passes through the origin! This single point $(0,0)$ serves as both the x-intercept and the y-intercept. It's a pretty significant point for our graph, showing us where the function crosses both axes. Knowing these intercepts helps anchor our graph and gives us specific points to plot, making it easier to connect the dots and visualize the function's path.

Analyzing Function Behavior Near Asymptotes

Alright, now for the nitty-gritty: understanding how our function behaves near those vertical asymptotes at $x = 1$ and $x = -1$, and how it approaches the horizontal asymptote $y=0$. This is where things get exciting, guys, because we're looking at limits without explicitly using limit notation (for now!). Let's consider the region to the right of $x=1$. If we pick an x-value slightly larger than 1, say $x = 1.1$, the numerator $2x$ will be positive (around 2.2). The denominator $x^2-1$ will be $(1.1)^2 - 1 = 1.21 - 1 = 0.21$, which is a small positive number. A positive number divided by a small positive number is a large positive number. So, as x approaches 1 from the right ($x o 1^+$), $f(x) o +\infty$. Now, let's look just to the left of $x=1$. Take $x = 0.9$. The numerator $2x$ is positive (around 1.8). The denominator $x^2-1$ is $(0.9)^2 - 1 = 0.81 - 1 = -0.19$, a small negative number. A positive number divided by a small negative number is a large negative number. So, as x approaches 1 from the left ($x o 1^-$), $f(x) o -\infty$. This tells us the graph shoots upwards on the right side of $x=1$ and downwards on the left side. We can do similar analysis for $x=-1$. To the right of $x=-1$ (e.g., $x = -0.9$), numerator is negative, denominator is negative (e.g., $(-0.9)^2 - 1 = 0.81 - 1 = -0.19$). Negative divided by negative is positive. So, as $x o -1^+$, $f(x) o +\infty$. To the left of $x=-1$ (e.g., $x = -1.1$), numerator is negative, denominator is positive (e.g., $(-1.1)^2 - 1 = 1.21 - 1 = 0.21$). Negative divided by positive is negative. So, as $x o -1^-$, $f(x) o -\infty$. These insights are crucial for sketching the correct graph, showing how the function behaves wildly near its vertical asymptotes. It helps us understand the curves and turns that define the function's shape.

Sketching the Graph and Identifying the Correct Representation

Okay, we've gathered all the crucial intel: vertical asymptotes at $x = 1$ and $x = -1$, a horizontal asymptote at $y = 0$, and intercepts at $(0, 0)$. We also know the function's behavior near the vertical asymptotes: it goes to $+\infty$ as $x o 1^+$ and $x o -1^+$, and to $-\infty$ as $x o 1^-$ and $x o -1^-$. Now, let's put it all together to visualize or sketch the graph. We can plot the intercepts and draw the asymptotes as dashed lines. Then, we use the behavior analysis to connect the pieces. For $x > 1$, the graph starts near $y=0$ (the horizontal asymptote), goes up towards $+\infty$ as it approaches $x=1$ from the right. For $-1 < x < 1$, the graph passes through $(0,0)$. As $x$ approaches $1$ from the left, it goes to $-\infty$. As $x$ approaches $-1$ from the right, it goes to $+\infty$. This suggests a curve that goes down towards $-\infty$ near $x=1$ and up towards $+\infty$ near $x=-1$, with $(0,0)$ being a peak or a valley between these two asymptotes. Since it goes to $-\infty$ as $x o 1^-$ and $+\infty$ as $x o -1^+$, and passes through $(0,0)$, this section of the graph will be a curve that arches upwards between $x=-1$ and $x=0$ and then arches downwards between $x=0$ and $x=1$. For $x < -1$, the graph starts near $y=0$ (the horizontal asymptote) for very negative x-values and goes down towards $-\infty$ as it approaches $x=-1$ from the left. So, when you look at the potential graphs, you're looking for a graph that has vertical asymptotes at $x=1$ and $x=-1$, a horizontal asymptote at $y=0$, passes through the origin $(0,0)$, and exhibits these specific behaviors in each of the three regions defined by the vertical asymptotes. The graph that accurately depicts all these characteristics is the one that represents f(x)=rac{2 x}{x^2-1}. It's all about matching the analytical findings with the visual representation. Keep practicing, and you'll be spotting these graphs in no time, guys!