Graphing Rational Function F(x)=(x^2-1)/(x^2-2x+1) Guide

by Axel Sørensen 57 views

Hey guys! Today, we're diving deep into understanding the graph of the rational function $f(x) = \frac{x^2 - 1}{x^2 - 2x + 1}$. Rational functions can seem a bit intimidating at first, but breaking them down step-by-step makes them super manageable. We’ll explore key features like holes, vertical asymptotes, and intercepts to get a clear picture of what this function looks like. By the end of this guide, you'll be able to confidently analyze similar functions and understand their graphical behavior. So, let’s jump right in and unravel the mysteries of this function!

Simplifying the Function

Before we start graphing, let’s simplify our function $f(x)$. Simplifying rational functions is crucial because it helps us identify any potential holes or discontinuities. The function we’re working with is $f(x) = \frac{x^2 - 1}{x^2 - 2x + 1}$.

First, we need to factor both the numerator and the denominator. The numerator, $x^2 - 1$, is a difference of squares, which factors into $(x - 1)(x + 1)$. The denominator, $x^2 - 2x + 1$, is a perfect square trinomial, which factors into $(x - 1)^2$. So, our function now looks like this:

f(x)=(x−1)(x+1)(x−1)2f(x) = \frac{(x - 1)(x + 1)}{(x - 1)^2}

Now, we can simplify further by canceling out the common factor of $(x - 1)$ from the numerator and the denominator. This gives us:

f(x)=x+1x−1f(x) = \frac{x + 1}{x - 1}

However, it’s super important to remember that we canceled out a factor of $(x - 1)$. This means there's a hole in the graph where $x = 1$. We'll come back to this later when we discuss holes in detail.

So, our simplified function is $f(x) = \frac{x + 1}{x - 1}$, but we must keep in mind the hole at $x = 1$. Simplifying helps us see the core behavior of the function, making it easier to identify asymptotes and intercepts. Understanding this simplification process is key to accurately graphing rational functions. Now that we have our simplified function, let’s move on to identifying the vertical asymptotes.

Identifying Vertical Asymptotes

Vertical asymptotes are super important features of rational functions. They indicate where the function's value shoots off to infinity or negative infinity. To find vertical asymptotes, we look at the denominator of the simplified function. Remember our simplified function is $f(x) = \frac{x + 1}{x - 1}$. The vertical asymptotes occur where the denominator equals zero. So, we need to solve the equation:

x−1=0x - 1 = 0

Adding 1 to both sides, we get:

x=1x = 1

So, we have a vertical asymptote at $x = 1$. This means that as $x$ approaches 1, the function's value will either increase without bound (approach positive infinity) or decrease without bound (approach negative infinity). However, remember that we had a hole at $x = 1$ due to the cancellation of the $(x - 1)$ factor earlier. This means that while the line $x = 1$ is a vertical asymptote for the simplified function, the original function has a hole at $x = 1$. The existence of the hole means that the function is undefined at $x = 1$, but the function still approaches vertical asymptotic behavior very close to this point, except for the single point of the hole.

Vertical asymptotes are crucial for understanding the behavior of a rational function because they show us the values where the function is undefined and where it experiences dramatic changes in value. To nail this down, let’s quickly recap. We look at the denominator of the simplified function, set it to zero, and solve for $x$. That gives us our vertical asymptote. Knowing where our vertical asymptotes are is a major step in sketching an accurate graph. Next up, we'll explore how to find horizontal asymptotes, which give us information about the function’s end behavior.

Determining Horizontal Asymptotes

Horizontal asymptotes tell us about the behavior of the function as $x$ approaches positive or negative infinity. In other words, they show us what value $f(x)$ approaches as $x$ gets extremely large or extremely small. There's a neat trick to finding horizontal asymptotes based on the degrees of the numerator and denominator of our rational function. Let’s look at our simplified function again:

f(x)=x+1x−1f(x) = \frac{x + 1}{x - 1}

Here, the degree of the numerator (the highest power of $x$) is 1, and the degree of the denominator is also 1. When the degrees are equal, we find the horizontal asymptote by dividing the leading coefficients of the numerator and the denominator. The leading coefficient is the number in front of the highest power of $x$.

In our case, the leading coefficient of the numerator is 1 (from the term $x$), and the leading coefficient of the denominator is also 1 (from the term $x$). So, we divide these leading coefficients:

11=1\frac{1}{1} = 1

This means we have a horizontal asymptote at $y = 1$. As $x$ approaches positive infinity or negative infinity, the function $f(x)$ will approach the value 1. This gives us a crucial piece of information about the long-term behavior of the function. It tells us that the graph will level off near the line $y = 1$ as we move far to the left or right on the x-axis.

To quickly recap, if the degrees of the numerator and denominator are the same, we divide the leading coefficients to find the horizontal asymptote. If the degree of the denominator is greater, the horizontal asymptote is $y = 0$. If the degree of the numerator is greater, there is no horizontal asymptote (but there might be a slant asymptote, which is a topic for another time!). Understanding horizontal asymptotes helps us complete the picture of how our function behaves over large intervals. Now that we’ve tackled horizontal asymptotes, let’s move on to finding the intercepts, which will give us specific points where the graph crosses the axes.

Finding Intercepts: Where the Graph Crosses the Axes

Intercepts are the points where the graph of the function crosses the x-axis and the y-axis. They are super helpful for sketching an accurate graph because they give us specific points to plot. Let's start with the y-intercept. The y-intercept is the point where the graph crosses the y-axis, which occurs when $x = 0$. To find it, we plug $x = 0$ into our simplified function:

f(x)=x+1x−1f(x) = \frac{x + 1}{x - 1}

f(0)=0+10−1=1−1=−1f(0) = \frac{0 + 1}{0 - 1} = \frac{1}{-1} = -1

So, the y-intercept is at the point $(0, -1)$. This means the graph crosses the y-axis at $y = -1$.

Next, let’s find the x-intercept(s). The x-intercepts are the points where the graph crosses the x-axis, which occur when $f(x) = 0$. To find them, we set the numerator of our simplified function equal to zero and solve for $x$:

x+1=0x + 1 = 0

Subtracting 1 from both sides, we get:

x=−1x = -1

So, the x-intercept is at the point $(-1, 0)$. This means the graph crosses the x-axis at $x = -1$.

Intercepts are really important because they give us anchor points for our graph. We know the graph passes through $(0, -1)$ and $(-1, 0)$. These points, combined with our knowledge of asymptotes, give us a good framework for sketching the function. To recap, the y-intercept is found by setting $x = 0$ in the function, and the x-intercept(s) are found by setting the numerator equal to zero. With our intercepts in hand, we’re getting closer to piecing together the complete picture. Now, let's talk about those pesky holes and how they affect our graph.

Understanding and Identifying Holes in the Graph

Holes are a unique feature of rational functions that can sometimes be a bit tricky. They occur when a factor is canceled out from both the numerator and the denominator during simplification. Remember when we simplified our function $f(x) = \frac{x^2 - 1}{x^2 - 2x + 1}$? We factored it to $ rac{(x - 1)(x + 1)}{(x - 1)^2}$ and then canceled out a factor of $(x - 1)$, giving us the simplified form $f(x) = \frac{x + 1}{x - 1}$.

The canceled factor, $(x - 1)$, tells us that there’s a hole in the graph at $x = 1$. To find the y-coordinate of the hole, we plug $x = 1$ into the simplified function (not the original, because that would give us an undefined result):

f(1)=1+11−1f(1) = \frac{1 + 1}{1 - 1}

Whoops! We can’t just plug it into the simplified function as is because we'll end up dividing by zero. Instead, we plug $x = 1$ into the simplified form before we simplified: $f(x) = \frac{x + 1}{x - 1}$. So,

f(1) = \frac{1 + 1}{1 - 1}$ is undefined. We made a mistake earlier! We should have plugged $x = 1$ into the *simplified* function $f(x) = \frac{x + 1}{x - 1}$ to find the y-coordinate of the hole: $f(1) = \frac{1 + 1}{1 - 1} = \frac{2}{0}

This is still undefined, which confirms that there's a discontinuity at $x = 1$. To find the y-coordinate of the hole, we need to take a step back and plug $x = 1$ into the simplified function before we simplified completely, meaning after we canceled out the common factor but still recognize the function's behavior:

f(x)=(x−1)(x+1)(x−1)(x−1)f(x) = \frac{(x - 1)(x + 1)}{(x - 1)(x - 1)}

We conceptually cancel one $(x-1)$ term but keep the simplified function’s form $f(x) = \frac{x + 1}{x - 1}$. Now plug in $x = 1$:

However, there’s an easier way! We plug $x = 1$ into the function after canceling out the common factor $(x-1)$ but before any further simplification:

f(x)=x+1x−1f(x) = \frac{x + 1}{x - 1}

But wait! We realize we can't directly substitute $x = 1$ into this simplified function either, because it still results in division by zero. This is a crucial point! We need to recognize that the hole exists because of the canceled factor, but the y-coordinate of the hole needs a different approach.

Let’s think about this differently. The hole occurs because the original function is undefined at $x = 1$, but the limit of the function as $x$ approaches 1 does exist. To find the y-coordinate of the hole, we need to consider the limit:

lim⁡x→1x+1x−1\lim_{x \to 1} \frac{x + 1}{x - 1}

However, simply substituting $x = 1$ leads to an indeterminate form. We need to analyze the function's behavior as $x$ approaches 1 from the left and from the right. This involves understanding that we have a vertical asymptote at $x = 1$, which complicates finding the exact y-coordinate of the hole directly.

Instead of a direct substitution, we should acknowledge that a hole at $x = 1$ in the original function means the function is undefined there, and the simplified form $f(x) = \frac{x + 1}{x - 1}$ (with the understanding that $x \neq 1$) represents the function everywhere except at that point. There's a discontinuity, but not a simple point we can calculate by plugging in $x = 1$.

The key takeaway here is that a hole indicates a point of discontinuity. The function behaves like the simplified form everywhere else, but we can't simply plug in the x-value of the hole into the simplified function to get a y-value. We've hit a bit of a snag here, and it highlights the complexity of dealing with holes in rational functions. Let's refocus on what we know and make sure we're on the right track.

Correct Approach for the Hole’s y-coordinate:

Okay, guys, let’s get this straight. The hole occurs at $x = 1$. To find the y-coordinate of the hole, we need to consider the function after we've canceled the common factor, but we can't directly substitute $x = 1$ because of the vertical asymptote.

What we need to do is look at the simplified function $f(x) = \frac{x + 1}{x - 1}$ and think about what happens as $x$ gets very close to 1, but isn't equal to 1. Since there's a vertical asymptote at $x = 1$, the function will approach either positive or negative infinity as $x$ gets close to 1. This means there isn’t a simple y-coordinate for the hole; instead, the function is undefined at that point, and we have to consider the behavior around the asymptote.

So, in practical terms, when graphing, you'd draw the simplified function $f(x) = \frac{x + 1}{x - 1}$ with a vertical asymptote at $x = 1$, and you wouldn't mark a specific hole point. The hole is represented by the fact that the function is undefined at $x = 1$.

Let's summarize how to identify holes:

  1. Factor the numerator and denominator of the rational function.
  2. Cancel any common factors.
  3. The values of $x$ that make the canceled factors equal to zero are the x-coordinates of the holes.
  4. Realize there's no simple y-coordinate to calculate due to the asymptotic behavior; the function is undefined at the hole's x-value.

Understanding holes is crucial for accurately graphing rational functions. They represent points where the function is undefined due to the cancellation of factors. Now that we’ve clarified this tricky concept, let's move on to putting it all together and sketching the graph.

Putting It All Together: Sketching the Graph

Alright, guys! We’ve done the hard work of figuring out all the key features of our function $f(x) = \frac{x^2 - 1}{x^2 - 2x + 1}$. Let’s recap what we know and then use that information to sketch the graph:

  • Simplified Function: $f(x) = \frac{x + 1}{x - 1}$
  • Vertical Asymptote: $x = 1$
  • Horizontal Asymptote: $y = 1$
  • y-intercept: $(0, -1)$
  • x-intercept: $(-1, 0)$
  • Hole: At $x = 1$, but no specific y-coordinate due to the vertical asymptote.

Now, let’s sketch the graph step-by-step:

  1. Draw the Asymptotes: First, draw the vertical asymptote at $x = 1$ as a dashed vertical line. This line indicates where the function is undefined and where it approaches infinity or negative infinity. Then, draw the horizontal asymptote at $y = 1$ as a dashed horizontal line. This line shows the value that the function approaches as $x$ goes to positive or negative infinity.

  2. Plot the Intercepts: Plot the y-intercept at $(0, -1)$ and the x-intercept at $(-1, 0)$. These points give us specific locations where the graph crosses the axes.

  3. Consider the Hole: We know there’s a hole at $x = 1$, but we don't have a specific y-coordinate to plot. This means that as the graph approaches $x = 1$, it will get very close to the vertical asymptote, but there won't be a defined point there.

  4. Sketch the Graph: Now, we can sketch the graph in the different regions defined by the asymptotes.

    • To the left of the vertical asymptote ($x < 1$): The graph passes through the x-intercept at $(-1, 0)$ and the y-intercept at $(0, -1)$. As $x$ approaches negative infinity, the graph approaches the horizontal asymptote $y = 1$. As $x$ approaches 1 from the left, the graph goes towards negative infinity.
    • To the right of the vertical asymptote ($x > 1$): The graph approaches the horizontal asymptote $y = 1$ as $x$ goes to positive infinity. As $x$ approaches 1 from the right, the graph goes towards positive infinity.

  5. Final Touches: Make sure your graph approaches the asymptotes but never crosses them (except that it can cross a horizontal asymptote). Indicate the hole by showing the function is undefined at $x = 1$.

By following these steps, we can create an accurate sketch of the rational function $f(x)$. Remember, the key is to break it down into smaller parts: simplify the function, identify asymptotes, find intercepts, and understand the behavior around any holes. This approach will help you tackle any rational function graphing problem with confidence!

Wrapping Up: Key Takeaways

Okay, guys, we’ve covered a lot in this comprehensive guide to graphing the rational function $f(x) = \frac{x^2 - 1}{x^2 - 2x + 1}$. Let’s quickly recap the key takeaways to make sure we’ve got it all nailed down:

  1. Simplification is Key: Always start by simplifying the rational function by factoring the numerator and denominator and canceling any common factors. This helps you identify holes and simplifies the function for further analysis.
  2. Vertical Asymptotes: Find vertical asymptotes by setting the denominator of the simplified function equal to zero and solving for $x$. These indicate where the function approaches infinity or negative infinity.
  3. Horizontal Asymptotes: Determine horizontal asymptotes by comparing the degrees of the numerator and denominator. If the degrees are equal, divide the leading coefficients. If the denominator’s degree is greater, the horizontal asymptote is $y = 0$.
  4. Intercepts: Find the y-intercept by setting $x = 0$ in the function. Find the x-intercepts by setting the numerator equal to zero and solving for $x$.
  5. Holes: Holes occur when factors are canceled out during simplification. The x-value of the hole is the value that makes the canceled factor equal to zero. There's no simple y-coordinate to calculate due to the function’s asymptotic behavior at the hole.
  6. Sketching the Graph: Draw asymptotes first, then plot intercepts. Sketch the graph in the regions defined by the asymptotes, making sure it approaches the asymptotes but doesn't cross them (except possibly the horizontal asymptote).

Graphing rational functions can seem daunting at first, but by breaking it down into these steps, you can confidently analyze and sketch these functions. Understanding these key features – asymptotes, intercepts, and holes – is essential for getting an accurate picture of the function’s behavior. Keep practicing, and you’ll become a pro at graphing rational functions in no time! Remember, math is like a puzzle; each piece of information fits together to create the whole picture. So, keep exploring, keep questioning, and most importantly, keep having fun with math!