Hey everyone, let's work through this example problem together. Here we want to graph the rational function \( f(x) = \frac{2x^2}{x^2 - 1} \). So let's go ahead and get started with step 1, which is to factor and find our domain. Here, when we factor, our numerator isn't going to factor anymore. It's simply just \(2x^2\), but my denominator does factor into \( (x+1)(x-1) \). From here, I can go ahead and find my domain by setting my denominator equal to zero. So I take my denominator, \(x+1=0\) and \(x-1=0\). Subtracting 1 from both sides here gives me \(x = -1\), and then adding 1 to both sides leaves you with \(x = 1\). Therefore, my domain restriction is \(x\) such that \(x\) cannot be equal to -1 or 1. Now it doesn't matter the order you write that in as long as your restriction has those two numbers. So let's move on to step number 2 and find the holes of our function. We want to set any common factors equal to zero, but we actually don't have any common factors here, so we don't have to worry about that. And since we also don't have any common factors, that tells us it's already in lowest terms.

Now let's go ahead and move on to step 3 and find our x intercepts and their behavior. We're going to set our numerator equal to zero here, which is \(2x^2\). Setting that equal to zero, if I divide both sides by 2, canceling that out, I get \(x^2 = 0\). Taking the square root of zero, I will simply end up with zero. You might have already noticed that before you solved, and if that's the case, don't worry about solving and just go straight to \(x = 0\). We want to check the multiplicity of this. Looking at our original function, since this is \(2x^2\), this comes from a factor that is squared. It occurs twice. It has a multiplicity of 2, so that is an even number, which tells us we're just going to touch the x-axis at this point and not fully cross it. So let's graph our x-intercept here. It's right at our origin at 0, and we're not sure if it's going to touch on the top of the x-axis or the bottom of the x-axis, so we'll wait to clarify that. Let's move on to finding our y intercepts by computing \(f(0)\), plugging 0 into our function, I get \( \frac{2 \times 0^2}{0^2 - 1} \). Since this gives me 0 on top of my fraction, I'm simply going to end up with 0, which you might have already noticed since our x-intercept is at that origin. That serves as both our x-intercept and our y-intercept here. So we can go ahead and move on to finding asymptotes here.

Let's start with our vertical asymptotes by setting our denominator equal to zero. Since we didn't cancel any factors, this is going to end up being the exact same as our domain. So just, again, setting that equal to zero, \(x-1=0\), and \(x+1=0\). Adding 1 to both sides gives me \(x = 1\), and then subtracting 1 from both sides gives me \(x = -1\). You don't have to redo that calculation if you already recognize that that's going to be the same as your domain restriction. But here, just to show you, there's my calculation. My 2 vertical asymptotes are at \(x = 1\) and \(x = -1\). So let's go ahead and plot that on our graph using a dashed line.

Next, let's find our horizontal asymptote. We're looking at the degrees of our numerator and our denominator here. Take another look at our function, \(f(x) = \frac{2x^2}{x^2 - 1}\). Both have a degree of 2, so the degree of my numerator is equal to the degree of my denominator. This tells me I need to divide my leading coefficients, looking back up at our function, I have \(2x^2\) in the numerator, so my leading coefficient is 2, and then I simply have \(x^2 - 1\) in the denominator. So it's just 1. This leaves me with a horizontal asymptote at \(y = 2\). Let's go ahead and plot that on our graph, again, using a dashed line.

Now that we have all of our asymptotes, we're going to break our graph up into intervals and then plot a point in each of them. Remember, when we're looking at our intervals, we want to go from something that we know, so we're going to start at our first-known piece of information, our vertical asymptote. Our very first interval is going to be from negative infinity to -1, from -1 to the next piece of information, I know is an x-intercept, so -1 to zero is my second interval, from zero to my next known piece of information, another vertical asymptote, is from 0 to 1. These are small intervals, but we still want to gather this information. Lastly, my final interval is going to be from 1 to infinity because I don't know anything else here. We want to choose a number in each interval to plot a point for.

So for my first interval from negative infinity to -1, I'm going to choose -3. From -1 to 0, this is a rather small interval, so I'll choose a fraction right in the middle, -1/2. I'm going to do the same from 0 to 1 and just choose positive 1/2. Then from 1 to infinity, I'll choose positive 2, giving me a good idea of what's happening on my graph. Now, we're going to plug these values into our function and find \(f(x)\) to get ordered pairs.

Starting with \(f(-3)\) and plugging -3 into our function. \(f(x) = \frac{2(-3)^2}{(-3)^2 - 1} \). This gives \( \frac{18}{8}\), which simplifies to approximately 2.25. So at -3, I get 2.25 as \(f(x)\). Next, I'll plug in -1/2, yielding: \( f(-1/2) = \frac{2(-1/2)^2}{(-1/2)^2 - 1} \) = \( \frac{2 \times \frac{1}{4}}{\frac{1}{4} - 1}\) = \( \frac{1/2}{-3/4} = \frac{-2}{3}\) or about -0.667. For +1/2, I get the same result, -0.667. Lastly, \(f(2)\) gives \( \frac{2 \times 4}{4-1} = \frac{8}{3}\), which is about 2.667.

Now let's plot these points on our graph and draw the graph accurately according to these calculations and the characteristics we've established. Thanks for following through, and I'll see you in the next one!