Discontinuities of rational functions | Mathematics III | High School Math | Khan Academy

So we have this function ( f(x) ) expressed as a rational expression here, or defined with a rational expression. We're told that each of the following values of ( x ) selects whether ( f ) has a zero, a vertical asymptote, or a removable discontinuity.

Before even looking at the choices, what I'm going to do—because you're not always going to have these choices here, sometimes you might just have to identify the vertical asymptotes, or the zeros, or the removable discontinuities—is I'm going to factor this out. Hopefully, I’m going to factor the numerator and the denominator to make things a little bit more clear.

We're going to think about what ( x ) values make the numerator and/or the denominator equal to 0. So, the numerator... can I factor that? Well, let's see. What two numbers, their product is negative 24 and they add to negative 2?

Let’s see, that would be negative 6 and positive 4. So we could write ( (x - 6)(x + 4) ). Did I do that right? Negative 6 times 4 is negative 24, and negative 6x plus 4x is negative 2x, yep, that's right.

All right, now, in the denominator, let’s see. ( 6 \times 4 = 24 ) and ( 6 + 4 = 10 ). So we could say ( 6(x + 4) ).

Let’s look at the numerator first: the numerator ( 0 ) if ( x = 6 ) or ( x = -4 ). The denominator ( 0 ) if ( x = -6 ) or ( x = -4 ).

So let’s think about it: can we simplify what we have up here a little bit? I mean, it might just be jumping out at you. You have ( (x + 4) ) divided by ( (x) ). Why can't we simplify this expression a little bit and just write this whole thing... write our original expression as being ( \frac{x - 6}{x + 6} ), where if we want to be algebraically equivalent, we have to constrain the domain where ( x ) cannot be equal to -4.

This is interesting because before, when ( x = -4 ), we had all this weird ( \frac{0}{0} ) stuff, but we were able to remove it, and we were able to make it just kind of this extra condition. If you were to graph it, it would show up as a little bit of a removable discontinuity, a little bit of a point discontinuity. It's just one point where the function is actually not defined.

So, in this situation—and this is typical—where there's a thing that you can factor out that was making the numerator and the denominator equal to 0, but you can factor that out, so it no longer does it. That's going to be a removable discontinuity. So, ( x = -4 ) is a removable discontinuity.

Once you've factored out all the things that would make it a removable discontinuity, then you can think about what's going to be a zero and what's going to be a vertical asymptote. If you've factored out everything that's common to the numerator and the denominator, if what's left makes the numerator equal to zero, well then it's going to make this whole expression equal to zero, and so you're dealing with a ( 0 ).

At ( x = 6 ), 6 would make the numerator equal to 0. ( 6 - 6 = 0 ). So there you’re dealing with a ( 0 ). To make the denominator equal to zero, you get ( x = -6 ); that would make the denominator equal zero. So that's a vertical asymptote.

It's called a vertical asymptote because as you approach this value, approaching negative six either from values smaller than negative six or larger than negative six, the denominator is going to become a very small positive number or a very small negative number. It's going to approach zero either from above or below, and so when you divide by that, you're going to get very large positive values or very large negative values.

So your graph does stuff like that. You have a vertical asymptote. It might either do something like that, or it might do something like this.

Let's do another one. All right, so same drill. Like before, always pause the video and see if you can work it out on your own.

Let me factor it. So let’s see, two numbers—the product is negative 32—they're going to have to have different signs: 8 and 4. Since they have to add up to positive 4, we want the larger one to be positive. So ( (x + 8)(x - 4) ), yep, I think that's right.

Then divided by this... ( 4 \times 4 = 16 ) and ( -4 + -4 = -8 ). So this is ( (x - 4)(x - 4) ).

Now this is really, really interesting because you might say, “Oh, I have ( (x - 4)(x - 4) ). Maybe ( x = 4 ) is a removable discontinuity.” It would have been unless you had... if you didn’t have this over here, because ( x = 4 ), even after you factor this, is still going to make the function not defined.

So this function is actually equivalent to ( \frac{x + 8}{x - 4} ). I don't have to put that extra little constraint like I did before. I don’t have to put this little constraint that describes this removable discontinuity because that constraint is still there after I've factored out, after I've canceled out this ( (x - 4) ) with this ( (x - 4) ).

And so here, we have... this is actually an algebraically equivalent expression to this one right over here. Now we can think about what the zeros, the vertical asymptotes, or the removable discontinuities are going to be.

Well, something that makes the numerator zero without doing it to the denominator is going to be a zero. ( x = -8 ) makes the numerator equal to zero without making the denominator equal to zero; it would make the denominator equal to negative 12.

So, you can literally just evaluate ( h(-8) ) and it would equal ( \frac{0}{-12} ), which is equal to 0.

So that's why we call it a ( 0 ). What about ( x = 4 )? Well, ( x = 4 ) is going to make the denominator only equal to 0. So that's going to give us a vertical asymptote.

And we're all done.