Graphs of rational functions: zeros | High School Math | Khan Academy

So we're told let ( F(x) = \frac{2x^2 - 18}{G(x)} ), where ( G(x) ) is a polynomial. Then they tell us which of the following is a possible graph of ( y = F(x) ). They give us four choices here, and like always, I encourage you to pause the video and see if you could give a go at it. Look at our ( F(x) ), and then think about which of these graphs actually match up or could match up to that ( F(x) ).

All right, now let's work through this together. They don't give us a lot of information; they don't tell us anything about the denominator of this rational expression, but they do tell us its numerator. Like we've seen before, it's useful to factor the numerator and see at what ( x ) values do interesting things happen. In particular, at what ( x ) values does the numerator equal zero?

So if we factor the numerator up here, we could rewrite ( F(x) ) as being equal to, let's see, we could factor out a two out of the numerator, so it's ( 2 \cdot (x^2 - 9) ), and that's all going to be over ( G(x) ). We don't know what the denominator is; we just know that it's a polynomial.

Let's see here, in the numerator ( x^2 - 9 ), you might recognize that as a difference of squares, so we can factor that further. We still have that original two, and it's going to be ( (x + 3)(x - 3) ), and all of that is still going to be over ( G(x) ).

The first thing that we might realize is, okay, when does our numerator equal zero? Well, when ( x = -3 ) or when ( x = 3 ). If ( x = -3 ), this expression is going to be zero; if ( x = 3 ), this expression is equal to zero. So you might just say, well, maybe we have zeros at plus or minus three. So maybe at ( F(-3) = 0 ) and ( F(3) = 0 ). Those values sure look like they make the numerator equal to zero.

Then we look at our choices. When we look at our choices, this choice A does seem to have a zero at positive 3, but it doesn't have one at negative 3; it has a vertical asymptote at -3. So that seems a little bit confusing. This choice B does have a zero at positive 3, but it has nothing going on here at negative 3. It defines negative 3; it doesn't even have a vertical asymptote there, so once again, this looks a little bit perplexing.

Choice C has a removable discontinuity at positive 3, and then it has a vertical asymptote at -2. So once again, this doesn't have anything interesting going on at ( x = -3 ). Still a little perplexing, and this one has zeros at positive 6 and negative 6. So none of the choices have zeros at both ( x = 3 ) and ( x = -3 ).

So what's going on? Well, what we need to realize is just because something makes the numerator equal to zero doesn't mean that it's definitely going to be a zero for that function. You might say, well, how can that be? Well, think about situations in which those values would also make the denominator equal to zero.

So let me write out some potential ( F(x) ) here. We just know that ( G(x) ) is a polynomial. So ( F(x) ) could be, we know the numerator ( 2 \cdot (x + 3)(x - 3) ) over, well, let's just say ( G(x) = x + 1 ). Well, in this situation, none of the values that make the numerator equal zero make the denominator equal zero, so this is a situation where you would have two zeros at ( x = 3 ) and ( x = -3 ).

So this would be the two zeros. Let's look at another situation; let's look at a situation where ( F(x) ) is equal to, we know the numerator ( (x + 3)(x - 3) ), and let's say that we do have, let's say that one of those ( x ) values, positive or negative 3, do make the denominator equal to zero.

So let's say ( G(x) = (x + 3)(x + 1) ). Well, you see here now since ( (x + 3) ) can is both in the numerator and the denominator, you could divide ( (x + 3) ) and they cancel out. Here, ( x = -3 ) would be a removable discontinuity. So this would have zero at ( x = 3 ) and a removable discontinuity at ( x = -3 ).

Those values that make the numerator equal to zero now could be a zero or it could represent a removable discontinuity. Here, I just picked a removable discontinuity to be at negative 3; it could be the other way around or it could be at both values if this was ( (x + 3)(x - 3) ) over ( (x + 3)(x - 3) ). Then you would have a removable discontinuity at both ( x = 3 ) and ( x = -3 ).

Then you could go even further: ( F(x) ) could look like this. It could be ( \frac{2 \cdot (x + 3)(x - 3)}{(x + 3)^2} \cdot (x + 1) ). So what's going to happen here? Even if you divide the numerator and the denominator by ( (x + 3) ), you're still going to have one ( (x + 3) ) left over in the denominator that could cancel with one of the ( (x + 3) )s, but you're still going to have an ( (x + 3) ).

So in this case, you would have a vertical asymptote at ( x = -3 ). So these particular examples that I just showed you demonstrated that any value that makes the numerator equal zero isn't necessarily a zero for the function. They could be zeros; they could be removable discontinuities, or they could be vertical asymptotes, but they would all occur at ( x = 3 ) or ( x = -3 ).

So with that lens, now let's look at the choices again. So choice A has a zero at ( x = 3 ), and it has a vertical asymptote at ( x = -3 ). So that's actually very consistent with this situation that I just described. So choice A actually is looking pretty good. Choice B has a zero at ( x = 3 ), but its vertical asymptote looks like it's at ( x = 2 ), and nothing interesting is happening at ( x = -3 , so we can rule that out.

As you look at choice C, you have a removable discontinuity at ( x = 3 ), which is completely possible. We've seen that situation where something that makes the numerator equals zero could be a removable discontinuity if you have that same expression in the denominator. But then the vertical asymptote isn't at ( x = -3 ); it's at ( x = -2 ), so that rules it out once again. Nothing interesting is happening at ( x = -3 ).

Here you have two zeros, but they're not at ( x = 3 ) or ( x = -3 ); they're at ( x = 6 ) and ( x = -6 ). So we can definitely rule that one out. So we should feel pretty good about choice A.