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. So here, when we factor, our numerator isn't going to factor anymore. It's simply just \( 2x^2 \), but my denominator does factor into \( (x+1)(x-1) \). So from here, I can go ahead and find my domain by setting my denominator equal to 0. So I take my denominator, \( x+1=0 \) and \( x-1=0 \). Now subtracting 1 from both sides here gives me \( x = -1 \). And then here, adding 1 to both sides leaves you with \( x=1 \). So here, my domain restriction is \( x \) such that \( x \neq -1 \) or \( x \neq 1 \). 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. Since we don't have any common factors here, we don't have to worry about holes, and we're already in lowest terms. So, let's move on to step 3 and find our \( x \)-intercepts and their behavior.
So we're going to set our numerator equal to 0 here, which is \( 2x^2 \). So setting that equal to 0, if I divide both sides by 2, canceling that out, I get \( x^2 = 0 \). And then if I take the square root of 0, I will simply end up with 0. Now 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 \( x = 0 \). Now 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.
Let's move on to finding our \( y \)-intercepts by computing \( f(0) \), plugging 0 into our function. So I get \( \frac{2 \times 0^2}{0^2 - 1} \). Since this gives me 0 on the top of my fraction, I'm simply just 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 0. Now since we didn't cancel any factors, this is going to end up being the exact same as our domain. So simply, again, setting that equal to 0, \( x-1 = 0 \), 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. Our 2 vertical asymptotes are at \( x = 1 \) and \( x = -1 \).
Let's go ahead and find our horizontal asymptote. Looking at the degree of our numerator and our denominator in the function \( \frac{2x^2}{x^2 - 1} \), both have a degree of 2. Since the degree of my numerator is equal to the degree of my denominator, I need to divide my leading coefficients. The numerator has 2 as the leading coefficient and the denominator has 1. This leaves me with a horizontal asymptote at \( y = 2 \).
Now that we have all of our asymptotes, we're going to go ahead and break our graph up into intervals and then plot a point in each of them. From here, we want to go ahead and actually plug these values into our original function and find \( f(x) \) in order to get ordered pairs. So let's go ahead and start with \( f(-3) \) and plug -3 into our original function. So \( f(-3) \). Looking back at my original function, I have \( 2 \times (-3)^2 \) over \( (-3)^2 - 1 \). Going ahead and multiplying that out, I know that \(-3^2\) is just going to be 9. So this is \( 2 \times 9 \) over \( 9 - 1 \), which gives me \( 18/8 \). Now simplifying that fraction, these are both divisible by 2, so this ends up giving me \( 9/4 \). Now if I were to plug that into a calculator, if it's easier for you to see this as a decimal, this is going to end up being 2.25, which will help us plot it on our graph.
So here at -3, I get 2.25 as \( f(x) \). Next, I plug in -1/2. So I have \( f(-\frac{1}{2}) = \frac{2 \times (-\frac{1}{2})^2}{(-\frac{1}{2})^2 - 1} \). This ends up being \( \frac{1/2}{-3/4} = -\frac{2}{3} \) or approximately -0.667 as a decimal. Continuing in this manner, we will plug in other values, calculate \( f(x) \), and plot these points to form the complete graph.
Let's go ahead and take all of these ordered pairs and plot them on our graph. We've made it through all of those calculations. Now we have one final step here, which is to connect everything and draw everything approaching our asymptote.
Looking at this side, I'm going to go ahead and have it approach my horizontal asymptote from there and then approach my vertical asymptote on either side. Remember that if you need to calculate extra points because you want a better idea, that's totally fine. Just pick some more values for \( x \), plug them into your function, and you're good to go. Now that we've finished graphing this, thanks for watching, and I'll see you in the next one.