WEBVTT
00:00:21.010 --> 00:00:24.744
Let's take a look at rational functions. We'll start with the definition.
00:00:25.179 --> 00:00:31.119
A rational function is a function of the form: f of x equals g of x over h of x,
00:00:31.314 --> 00:00:37.239
where g and h are polynomial functions such that the one in the denominator is not equal to 0.
00:00:39.362 --> 00:00:43.430
There's a special case of rational functions where h of x equals c,
00:00:43.755 --> 00:00:49.339
c being a non-zero real number. We'll consider those just to be polynomial functions
00:00:49.401 --> 00:00:52.046
and not treat them as rational functions.
00:00:53.622 --> 00:00:57.752
Two important features of every graph of every function we've ever studied
00:00:57.839 --> 00:01:02.541
are domain and intercepts. The domain for a rational function
00:01:03.741 --> 00:01:09.062
is all real numbers except for any which make the denominator, h of x, equal to 0.
00:01:12.086 --> 00:01:15.102
The intercepts are found just as we always find intercepts.
00:01:15.286 --> 00:01:20.963
To find the y-intercept, if it exists, let x equal 0 and evaluate the function.
00:01:22.263 --> 00:01:26.656
If there are any x-intercepts, you find them by solving f of x equals 0.
00:01:27.214 --> 00:01:31.623
We should note that rational functions may or may not have a y-intercept.
00:01:32.084 --> 00:01:33.770
If there is one, there is only one.
00:01:34.337 --> 00:01:37.369
x-intercepts may or may not exist as well,
00:01:37.632 --> 00:01:42.272
and a rational function may have zero, one, or many x-intercepts.
00:01:45.792 --> 00:01:48.073
In this example, we're given a rational function,
00:01:48.089 --> 00:01:51.353
and we're asked to find its domain and any intercepts.
00:01:52.598 --> 00:01:55.638
Recall that the domain of a rational function is all real numbers
00:01:55.648 --> 00:01:58.331
except any that make the denominator equal to 0.
00:01:58.685 --> 00:02:06.676
Let's start by finding any real numbers that make our denominator 0.
00:02:08.504 --> 00:02:16.365 , if you factor.
00:02:18.216 --> 00:02:22.934
Solving this, we see that x equals 15 and x equals negative 15
00:02:23.395 --> 00:02:26.496
are the two values that make our denominator equal to 0.
00:02:26.705 --> 00:02:29.211
We'll need to exclude these from our domain.
00:02:33.148 --> 00:02:36.961
We're usually asked to write our domains in interval notation, so let's do that.
00:02:37.187 --> 00:02:41.922
We want the set of all real numbers less than negative 15,
00:02:43.468 --> 00:02:50.429
between negative 15 and positive 15, as well as those that are greater than 15
00:02:51.836 --> 00:02:53.789
Here's the domain of our function.
00:02:55.367 --> 00:03:02.676
Now let's talk about intercepts. First, we're going to find the y-intercept.
00:03:04.696 --> 00:03:09.491
To do this, we let x equal 0, so we need to evaluate f of 0.
00:03:11.574 --> 00:03:19.756
That's going to give me 0 plus 1 over 0 squared minus 225.
00:03:21.941 --> 00:03:30.602
Simplifying, we get negative 1 over 225. That is our y-intercept.
00:03:30.904 --> 00:03:40.452
Now, the x-intercepts: if there are any, we find them by letting y equal 0.
00:03:41.196 --> 00:03:49.890
y is f of x, so we need to solve 0 equals x plus 1 over x squared minus 225.
00:03:50.668 --> 00:03:54.906
The first step in solving this is we want to multiply both sides
00:03:55.076 --> 00:04:03.541
by the denominator, x squared minus 225. So I end up with 0 equals x plus 1.
00:04:04.402 --> 00:04:08.099
x equals negative 1 is our only x-intercept.
00:04:08.947 --> 00:04:12.242
Notice that you might see a shortcut you might want to take.
00:04:14.311 --> 00:04:18.364
Since the first thing we did was eliminate the denominator,
00:04:18.510 --> 00:04:25.223
it turns out that the x-intercepts are always the solutions of 0 equal to your numerator
00:04:26.180 --> 00:04:29.041
as long as there aren't any common factors.
00:04:32.828 --> 00:04:36.819
Let's look at another example. Again, we're given a rational function,
00:04:36.850 --> 00:04:39.944
and we're asked to find its domain and any intercepts.
00:04:40.862 --> 00:04:44.466
To find the domain, we start by finding any values
00:04:44.658 --> 00:04:48.370
that will make our denominator equal to 0 so that we can exclude them.
00:04:49.360 --> 00:04:51.650
We factor out the common factor of x.
00:04:56.393 --> 00:05:00.740
x squared plus 4 can't be factored any further over the set of real numbers,
00:05:00.773 --> 00:05:05.692
so the only number that needs to be excluded from our domain is x equals 0.
00:05:10.094 --> 00:05:15.237
Our domain in interval notation is all real numbers less than 0
00:05:15.498 --> 00:05:17.902
along with all real numbers greater than 0.
00:05:20.893 --> 00:05:28.239
Now let's look at the intercepts. We've just excluded x equals 0 from our domain,
00:05:28.786 --> 00:05:43.550
so when I go to find the y-intercept, there can't be one since x can't be equal to 0,
00:05:46.818 --> 00:05:56.049
but there may still be some x-intercepts. Let's look at our function.
00:05:57.635 --> 00:06:06.513
I want to remove any common factors before I look for my x-intercepts,
00:06:06.559 --> 00:06:10.456
so both the numerator and denominator have a common factor of x.
00:06:17.622 --> 00:06:24.366
Dividing those out, I'm left with x plus 5 over x squared plus 4.
00:06:26.271 --> 00:06:31.809
The x-intercepts will be any values of x that make our numerator equal to 0,
00:06:38.588 --> 00:06:41.434
so x equals negative 5 is our x-intercept.
00:06:45.703 --> 00:06:48.595
Many rational functions have vertical asymptotes.
00:06:49.152 --> 00:06:51.611
Let's talk about what a vertical asymptote is.
00:06:52.218 --> 00:06:56.348
I've drawn you several examples. The vertical line x equals a
00:06:56.456 --> 00:07:00.911
is a vertical asymptote of a rational function or any function
00:07:00.934 --> 00:07:05.528
if at least one of the following occurs. In this case,
00:07:05.682 --> 00:07:10.168
the function values, or the y-values, are getting larger and larger as you get close to a
00:07:10.936 --> 00:07:16.706
from the right-hand side. We say that the function values are approaching infinity
00:07:16.731 --> 00:07:19.438
as x approaches a from the right.
00:07:21.451 --> 00:07:26.246
In this case, our function values are getting smaller and smaller as we approach a
00:07:26.385 --> 00:07:33.160
from the right again. We say that our function values are approaching negative infinity
00:07:33.363 --> 00:07:35.668
as x approaches a from the right.
00:07:36.549 --> 00:07:42.751
In this case, we're approaching a from the left,
00:07:43.750 --> 00:07:46.410
and again our function values here are rising,
00:07:46.579 --> 00:07:52.705
so we say that the function values are approaching infinity as x approaches a from the left.
00:07:53.788 --> 00:07:59.140
Finally, in this picture, our function values are getting smaller and smaller
00:07:59.186 --> 00:08:02.079
as we approach a again from the left,
00:08:03.126 --> 00:08:07.682
and we say that f of x approaches negative infinity as x approaches a from the left.
00:08:08.382 --> 00:08:12.543
If any one or more of these occur at x equals a,
00:08:12.819 --> 00:08:14.853
then that is a vertical asymptote for the graph.
00:08:16.008 --> 00:08:22.732
A rational function in particular will have, as long as g and h have no common factors,
00:08:23.640 --> 00:08:28.992
will have a vertical asymptote at x equals a whenever h of a equals 0.
00:08:32.495 --> 00:08:34.815
Let's talk more about vertical asymptotes.
00:08:35.837 --> 00:08:39.851
Here we're given a rational function, and we're asked to find all of its vertical asymptotes
00:08:40.178 --> 00:08:42.537
and then create a rough sketch
00:08:42.544 --> 00:08:46.752
to describe what the graph looks like near each of the asymptotes we find.
00:08:48.243 --> 00:08:54.398
Recall that a rational function will have a vertical asymptote at any value a
00:08:54.737 --> 00:08:59.819
that makes the denominator 0 as long as it doesn't also make the numerator 0,
00:08:59.900 --> 00:09:02.514
so we need to remove common factors before we begin.
00:09:05.128 --> 00:09:08.113
In the numerator, I can factor out the common factor x.
00:09:12.737 --> 00:09:15.065
The denominator's a trinomial that can be factored.
00:09:18.863 --> 00:09:21.384
We're going to use factors of 3, which are 1 and 3,
00:09:22.348 --> 00:09:25.983
and we need that 3 to be negative to make this work out.
00:09:29.102 --> 00:09:31.399
Okay, so we do have some common factors.
00:09:33.220 --> 00:09:38.834
Dividing out the common factor, we're left with x over x plus 1.
00:09:40.970 --> 00:09:44.925
The only number that makes our denominator equal to 0 is negative 1,
00:09:45.038 --> 00:09:58.943
so our only vertical asymptote is going to be x equals negative 1.
00:10:01.217 --> 00:10:05.084
Alright, now you probably are thinking, but what about the 3?
00:10:05.407 --> 00:10:10.178
3 is not in the domain because it makes our original function undefined.
00:10:10.909 --> 00:10:14.575
Well, that's a special case that will come up a little bit later in our video.
00:10:18.650 --> 00:10:23.110
We've just determined that this rational function has one vertical asymptote
00:10:23.111 --> 00:10:25.809
at x equals negative 1. Let's sketch that.
00:10:32.992 --> 00:10:36.055
Our next task is to determine how does the function behave
00:10:36.074 --> 00:10:41.049
as we get very close to this asymptote on either side. We will use test values.
00:10:41.452 --> 00:10:47.701
Start by picking numbers just a little bit to the left of negative 1.
00:10:47.941 --> 00:10:52.179
You want to be close, so don't pick negative 2 or something smaller than that.
00:10:52.804 --> 00:10:54.815
Let's start with say negative 1.3.
00:10:55.395 --> 00:11:00.909
Substitute negative 1.3 in place of the x in your reduced rational function.
00:11:03.797 --> 00:11:07.664
This is the first step. Simplify the denominator,
00:11:09.344 --> 00:11:15.001
so f of negative 1.3 is 13/3. That's a little bit more than 4.
00:11:16.740 --> 00:11:20.700
We need to go a little bit closer to try to get a better handle on the behavior,
00:11:21.330 --> 00:11:27.823
so now choose x to be negative 1.2. Substitute into our reduced rational function.
00:11:29.339 --> 00:11:34.768
Your first step will be negative 1.2 over negative 1.2 plus 1.
00:11:36.687 --> 00:11:44.792
Subtract in the denominator, and this value is 6. f of negative 1.2 is 6.
00:11:46.430 --> 00:11:50.336
Getting a little bit closer, let x be negative 1.1.
00:11:50.348 --> 00:11:55.522
Again, substitute into the reduced rational function.
00:11:58.127 --> 00:12:00.998
Working that out as we did before, this one is 11,
00:12:01.400 --> 00:12:03.589
so now we look to see is there a pattern?
00:12:04.655 --> 00:12:08.932
This is a litter bit more than 4. This one's bigger at 6. This one's bigger at 11.
00:12:09.240 --> 00:12:14.677
If we're not sure of the pattern, we could pick more test values closer and closer.
00:12:15.723 --> 00:12:19.304
This certainly seems reasonable that our values are getting larger and larger
00:12:19.309 --> 00:12:22.511
as our x gets closer and closer to negative 1.
00:12:23.174 --> 00:12:31.666
Our function, our y-values, are getting bigger as we get closer to negative 1 from the lefthand side.
00:12:36.002 --> 00:12:42.200
Now what about from the right? Again, we want to choose values close to negative 1,
00:12:42.317 --> 00:12:46.563
so we can't start way over here at 0 or 1. We need to be much closer in.
00:12:47.217 --> 00:12:52.006
We'll start with say negative 0.7. Let x be negative 0.7.
00:12:52.230 --> 00:12:55.912
Substitute that for x in our reduced rational function.
00:12:57.182 --> 00:13:01.784
That's going to be negative 0.7 over negative 0.7 plus 1.
00:13:03.287 --> 00:13:07.556
Simplify the denominator. You get negative 7/3.
00:13:09.509 --> 00:13:12.139
Repeat that for negative 0.8,
00:13:14.004 --> 00:13:19.557
substituting again in our reduced rational function and simplifying to get negative 4.
00:13:21.214 --> 00:13:28.801
Then finally, we'll use negative 0.9. Substitute in the reduced rational function and simplify.
00:13:30.429 --> 00:13:34.343
Now let's look for a pattern coming from the right side of x equals negative 1.
00:13:35.404 --> 00:13:40.501
This is negative 3 and 1/3, negative 4, negative 9,
00:13:40.934 --> 00:13:44.453
so it seems reasonable that our values are getting smaller and smaller
00:13:45.085 --> 00:13:49.153
as x gets closer and closer to negative 1 from the right.
00:13:50.646 --> 00:13:54.698
Okay, so these are our y-values, our function values, getting smaller and smaller
00:13:55.450 --> 00:13:57.855
as x approaches negative 1 from the right.
00:13:59.002 --> 00:14:03.960
This graph describes the behavior of our function f very close to the asymptote,
00:14:04.791 --> 00:14:07.691
and I'd like to point out that if you're not convinced at this point,
00:14:07.701 --> 00:14:10.176
you can always pick more and more values.
00:14:10.474 --> 00:14:14.024
The correct number to choose is what shows you the pattern.
00:14:17.642 --> 00:14:21.470
Now let's take a look at horizontal asymptotes. Here's the definition:
00:14:22.065 --> 00:14:25.978
If the values of f of x approach a fixed real number H
00:14:26.433 --> 00:14:30.316
as x approaches infinity or as x approaches negative infinity,
00:14:31.681 --> 00:14:36.043
then y equals h is a horizontal asymptote of f of x.
00:14:37.312 --> 00:14:40.313
Make sure you understand x approaches infinity
00:14:40.388 --> 00:14:43.644
means that your x-values are getting larger and larger and larger,
00:14:43.869 --> 00:14:47.109
so we will look towards the right side of our graph.
00:14:47.202 --> 00:14:49.314
As x gets larger and larger and larger,
00:14:49.476 --> 00:14:54.804
notice that these function values are getting closer and closer to the line y equals H.
00:14:56.846 --> 00:14:58.965
As x approaches negative infinity
00:14:59.117 --> 00:15:02.373
means that our x-values are getting smaller and smaller and smaller,
00:15:02.391 --> 00:15:04.503
so we're looking towards the left side of our graph.
00:15:04.552 --> 00:15:08.880
As the x-values approach negative infinity, notice that these y-values
00:15:08.895 --> 00:15:14.053
are also getting closer and closer to the horizontal line y equals H.
00:15:16.498 --> 00:15:17.805
Here's a second example.
00:15:19.484 --> 00:15:23.204
As the x-values get larger and larger as they approach infinity,
00:15:23.216 --> 00:15:25.737
the function values are getting closer and closer to H.
00:15:26.024 --> 00:15:28.971
The graph is getting closer and closer to the line y equals H,
00:15:28.986 --> 00:15:31.075
so we have another horizontal asymptote.
00:15:33.674 --> 00:15:39.412
As x approaches negative infinity, again the function values are getting closer and closer to H,
00:15:40.813 --> 00:15:44.340
and the graph itself is getting closer and closer to the line y equals H.
00:15:45.921 --> 00:15:49.842
The graph on the left is a little bit different. It still has a horizontal asymptote,
00:15:49.856 --> 00:15:55.296
but notice that you only need one side to look like an asymptote. Right?
00:15:55.447 --> 00:16:00.227
The y-values get closer and closer to y equals H as x approaches infinity.
00:16:00.513 --> 00:16:02.571
That's enough to make this a horizontal asymptote.
00:16:03.481 --> 00:16:09.143
Notice that the graph actually crosses over that horizontal asymptote over here somewhere.
00:16:09.176 --> 00:16:14.072
That's okay. A function may or may not cross its horizontal asymptote.
00:16:14.095 --> 00:16:17.513
It might cross it many times. That's okay.
00:16:18.623 --> 00:16:21.160
A function never crosses its vertical asymptotes.
00:16:24.778 --> 00:16:27.771
Let's talk more about horizontal asymptotes.
00:16:28.357 --> 00:16:34.475
Here's a rational function of the form g of x over h of x where g and h are polynomials,
00:16:35.252 --> 00:16:42.538
and we'll write them out into this form where we can see all of the terms of the polynomials.
00:16:44.057 --> 00:16:50.043
The a sub n, a sub of n minus 1, and so on are just any real numbers.
00:16:50.643 --> 00:16:54.278
n is the degree of the polynomial g in the numerator,
00:16:54.808 --> 00:16:59.255
and m is the degree of the polynomial h in the denominator.
00:17:01.475 --> 00:17:04.916
A rational function can have many vertical asymptotes,
00:17:05.276 --> 00:17:09.081
but a rational function can have at most one horizontal asymptote.
00:17:10.606 --> 00:17:15.486
Whether or not a horizontal asymptote exists, and where it is, if it does exist,
00:17:15.553 --> 00:17:18.206
depends on the relationship between m and n.
00:17:24.569 --> 00:17:43.696
If m is greater than n, then y equals 0 is the horizontal asymptote for the function.
00:17:44.730 --> 00:18:06.300
If m equals n, then y equals the ratio of the leading coefficients will be your horizontal asymptote.
00:18:07.481 --> 00:18:30.875
Finally, if m is less than n, then the function has no horizontal asymptote.
00:18:31.627 --> 00:18:33.947
Next, we'll look at some specific examples.
00:18:37.320 --> 00:18:40.181
In this example, we'll consider three rational functions
00:18:40.351 --> 00:18:43.723
and look for their horizontal asymptote if it exists.
00:18:46.337 --> 00:18:48.613
Remember that the relationship between
00:18:48.644 --> 00:18:53.640
the power of the numerator and the power of the demoninator, n and m,
00:18:54.009 --> 00:18:58.325
will help us to determine if there is a horizontal asymptote and where it is.
00:18:58.878 --> 00:19:06.717
In this first one, the degree of the numerator is 3. The degree of the denominator is 1,
00:19:07.054 --> 00:19:13.248
so since m is less than n, there is no horizontal asymptote.
00:19:22.219 --> 00:19:28.391
In our second example, the degree of the numerator is 1. The degree of the denominator is 3,
00:19:30.547 --> 00:19:50.448
so m is greater than n. The horizontal asymptote will be y equals 0.
00:19:52.313 --> 00:19:56.443
Finally, let's look at this one. Please don't be tricked.
00:19:57.156 --> 00:20:02.168
The degree of the numerator is 2 and so is the degree of the denominator.
00:20:04.656 --> 00:20:20.681
Since m equals n, the horizontal asymptote is y equals the ratio of the leading coefficients,
00:20:20.786 --> 00:20:29.173
so that would be positive 2 over negative 17.
00:20:32.275 --> 00:20:36.791
Let's look at two special rational functions, their graphs, and some of their properties.
00:20:37.087 --> 00:20:45.891
The first function is f of x equals 1 over x. 0 makes this denominator equal to 0,
00:20:45.905 --> 00:20:52.283
so it has to be excluded from the domain, so our domain here is all real numbers less than 0
00:20:52.314 --> 00:20:56.016
along with all real numbers greater than 0 written in interval notation.
00:20:57.649 --> 00:21:01.802
The range is the same thing: y can't equal 0.
00:21:02.830 --> 00:21:06.426
It has no intercepts, and it has two asymptotes.
00:21:06.434 --> 00:21:13.458
The vertical asymptote is at x equals 0. The x equals 0 is the y-axis.
00:21:15.958 --> 00:21:22.639
It has another asymptote: the horizontal asymptote at y equals 0. y equals 0 is the x-axis.
00:21:27.136 --> 00:21:30.346
You notice when you look at this that it is symmetric about the origin,
00:21:30.979 --> 00:21:36.284
so this is an odd function. Because it's an odd function,
00:21:36.287 --> 00:21:40.657
it's true that f of negative x is equal to negative f of x.
00:21:44.396 --> 00:21:47.799
Now let's look at f of x equals 1 over x squared.
00:21:48.233 --> 00:21:52.332
Again, x cannot equal 0, so we exclude 0 from the domain.
00:21:55.056 --> 00:22:01.074
Now look at the range. x can be any positive real number, and that's it,
00:22:01.214 --> 00:22:07.765
so the range will be 0 to infinity with a parenthesis on the 0 since it's not included in the range.
00:22:09.942 --> 00:22:16.625
The graph has no intercepts. It also has a vertical asymptote: x equals 0, the y-axis.
00:22:18.806 --> 00:22:22.890
It has a horizontal asymptote: y equals 0, which is the x-axis.
00:22:27.260 --> 00:22:31.414
This graph has a different kind of symmetry. It's symmetric with respect to the y-axis.
00:22:31.546 --> 00:22:38.159
This is an even function, and for even functions f of negative x is the same as f of x.
00:22:39.292 --> 00:22:44.760
We will use these two functions and transformations to graph some other rational functions.
00:22:48.597 --> 00:22:52.425
In this example, we're asked to use those two special rational functions
00:22:53.189 --> 00:22:56.430
to graph rational functions using transformations.
00:22:56.933 --> 00:23:01.256
Let's look at the first one: f of x equals negative 1 over x plus 4.
00:23:01.705 --> 00:23:03.909
What type of transformations will you need to do?
00:23:05.885 --> 00:23:13.573
Correct. The negative in front of 1 over x tells me that there's a reflection across the x-axis,
00:23:13.798 --> 00:23:19.165
and the plus 4 sort of outside tells me the graph should be shifted up 4 units.
00:23:19.876 --> 00:23:28.351
We'll start with the reflection across the x-axis.
00:23:29.669 --> 00:23:32.670
This point originally is at (1, 1).
00:23:33.082 --> 00:23:38.836
When we reflect that across the x-axis, the new y-coordinate will be negative 1.
00:23:39.057 --> 00:23:40.364
The x doesn't change.
00:23:41.180 --> 00:23:47.545
The behavior of the graph is still that it has asymptotes at the x-axis and the y-axis.
00:23:49.894 --> 00:23:52.833
This point, the y-coordinate right now is negative 1.
00:23:52.847 --> 00:23:58.338
When we reflect it across the x-axis, the new coordinates will be (-1, 1).
00:24:00.843 --> 00:24:06.450
Note that reflecting the graph across the x-axis doesn't affect the asymptotes at all.
00:24:13.332 --> 00:24:19.821
Okay, now we'll take our reflected graph and shift everything up 4 units.
00:24:21.346 --> 00:24:24.146
As soon as we start moving up or down,
00:24:24.174 --> 00:24:27.105
then we're moving our horizontal asymptote that amount as well,
00:24:27.178 --> 00:24:34.038
so instead of y equals 0, our asymptote will now be at y equals 4.
00:24:41.568 --> 00:24:53.718
Now we can move the points up 4 units. Sketch the graph of that little section,
00:24:54.439 --> 00:24:58.281
and then we have one more point to move up 4 units here.
00:24:59.693 --> 00:25:04.171
Instead of the y-coordinate here at 1, we move that up 4 units. It'll now be at 5.
00:25:05.031 --> 00:25:09.602
We draw that section of the graph using the asymptotes to guide us.
00:25:10.508 --> 00:25:14.074
This is the graph of y equals negative 1 over x plus 4.
00:25:15.032 --> 00:25:18.984
The vertical asymptote is still x equals 0.
00:25:20.601 --> 00:25:24.205
The horizontal asymptote is y equals 4.
00:25:25.926 --> 00:25:29.530
There are no y-intercepts. There is an x-intercept.
00:25:29.638 --> 00:25:33.235
If we wanted to find its value, then we know techniques for doing that.
00:25:37.094 --> 00:25:42.546
Now let's look at this function. This is 3 over x plus 2 squared.
00:25:44.476 --> 00:25:46.672
What sort of transformations will we need to do?
00:25:49.772 --> 00:25:53.431
The plus 2 inside sort of with the x
00:25:53.506 --> 00:25:58.974
tells me that I need to shift the graph to the left 2 units,
00:26:00.070 --> 00:26:04.517
and the 3 tells me the graph will be stretched vertically by a factor of 3.
00:26:05.135 --> 00:26:16.424
We need to do the horizontal shift first.
00:26:18.229 --> 00:26:22.854
If we're moving the graph left or right, that's going to affect our vertical asymptote.
00:26:23.037 --> 00:26:32.041
In this case, we're going to move our vertical asymptote from 0 to the left 2 units,
00:26:33.508 --> 00:26:36.276
so our vertical asymptote is x equals negative 2.
00:26:36.809 --> 00:26:51.300
Then we can move the points and the pieces of our graph to the left 2.
00:26:52.696 --> 00:26:55.975
We've changed our vertical asymptote with this transformation.
00:26:56.006 --> 00:26:58.559
Notice that we have not changed the horizontal asymptote.
00:27:01.349 --> 00:27:06.824
Alright, there's one last transformation.
00:27:07.924 --> 00:27:11.737
Vertical stretching changes the y-coordinates of the points.
00:27:12.111 --> 00:27:15.777
We can take the y-coordinates of this graph.
00:27:16.278 --> 00:27:20.671
We're going to multiply them by 3 to obtain the y-coordinates of our new graph.
00:27:21.577 --> 00:27:28.677
If we look at this point to start with, the y-coordinate is 1. 1 times 3 is 3.
00:27:33.276 --> 00:27:35.334
We should go ahead and draw that vertical asymptote.
00:27:35.341 --> 00:27:38.157
Stretching does not affect the vertical asymptote.
00:27:40.906 --> 00:27:43.226
It also doesn't affect the horizontal asymptote,
00:27:43.231 --> 00:27:48.258
so, when we draw this piece of the graph, we're still approaching the x-axis
00:27:48.262 --> 00:27:51.866
from the top as x goes towards infinity.
00:27:54.133 --> 00:27:57.258
Then over here, the y-coordinate is 1.
00:27:57.974 --> 00:28:01.353
Multiply that by 3. The new y-coordinate will be 3.
00:28:12.969 --> 00:28:16.689
Okay, so for this graph, to summarize:
00:28:16.699 --> 00:28:20.180
our vertical asymptote is now at x equals negative 2.
00:28:21.162 --> 00:28:24.193
The horizontal asymptote is still y equals 0.
00:28:26.728 --> 00:28:31.198
It does have a y-intercept. It might not really be at 1.
00:28:32.113 --> 00:28:34.951
Let x equal 0, and you solve for your y-intercept.
00:28:35.716 --> 00:28:37.410
It doesn't have any x-intercepts.
00:28:41.185 --> 00:28:43.962
Let's reconsider a function from earlier in this video:
00:28:44.654 --> 00:28:49.704
f of x equals x squared minus 3x over x squared minus 2x minus 3.
00:28:50.376 --> 00:28:52.534
The first thing I want to do is find its domain.
00:28:53.385 --> 00:28:56.100
For a rational function, the domain will be all real numbers
00:28:56.106 --> 00:28:59.122
except any that make the denominator equal to 0.
00:28:59.342 --> 00:29:17.967
We'll start by factoring the denominator and looking for the numbers that make it 0.
00:29:19.330 --> 00:29:25.517
We see after factoring that x equals 3 or x equals negative 1 will make this denominator 0.
00:29:25.700 --> 00:29:33.499
Those are the values we need to exclude from the domain.
00:29:34.239 --> 00:29:41.353
The domain will be all real numbers less than negative 1,
00:29:41.478 --> 00:29:47.936
between negative 1 and positive 3, and greater than 3.
00:29:50.434 --> 00:29:52.646
Now let's talk about vertical asymptotes.
00:29:54.113 --> 00:29:58.150
A rational function has a vertical asymptote at x equals a
00:29:59.140 --> 00:30:02.914
if a makes the denominator 0 but not the numerator.
00:30:04.458 --> 00:30:14.609
We need to remove any common factors. Factor the numerator.
00:30:16.868 --> 00:30:19.977
We see that there is a common factor of x minus 3.
00:30:21.688 --> 00:30:25.470
Reducing to lowest terms, we're left with x over x plus 1.
00:30:27.478 --> 00:30:31.584
We know that there's a vertical asymptote now at x equals negative 1,
00:30:38.622 --> 00:30:40.772
but what about x equals 3?
00:30:41.046 --> 00:30:44.310
It's not in the domain, but there is no vertical asymptote.
00:30:44.826 --> 00:31:01.639
As it turns out, there's a hole in the graph at x equals 3.
00:31:02.406 --> 00:31:18.007
We can find the coordinates of this hole using our reduced rational function.
00:31:20.636 --> 00:31:36.630
When x is 3, we would get 3/4, so the hole is going to occur when x is 3 with a y-value of 3/4.
00:31:39.450 --> 00:31:45.158
What we'd like to do next is see this on the graph and how it affects the graph as a whole.
00:31:48.164 --> 00:31:50.577
Another piece of information we need to be able to draw the graph
00:31:50.601 --> 00:31:55.636
will be the horizontal asymptote. We need to look at the reduced rational function,
00:31:55.987 --> 00:32:00.998
and we see that the degree of the numerator and the degree of the denominator are both 1.
00:32:01.638 --> 00:32:05.961
The horizontal asymptote will be the ratio of the leading coefficients.
00:32:09.172 --> 00:32:11.198
In this case, that's y equals 1.
00:32:14.487 --> 00:32:16.668
Now let's see what we can do with the graph.
00:32:21.031 --> 00:32:25.997
The horizontal asymptote is y equals 1, so we'll start there.
00:32:27.040 --> 00:32:30.582
The vertical asymptote we found is at x equals negative 1.
00:32:38.063 --> 00:32:41.420
So far, the only other thing we know is there's a hole in the graph
00:32:41.683 --> 00:32:44.398
at this point, so I'll go ahead and start with that.
00:32:49.159 --> 00:32:52.384
This was 1, so 3/4 is about right here.
00:32:54.988 --> 00:32:59.134
How am I going to finish the graph? At this point, I just need to use some test values.
00:33:02.506 --> 00:33:08.324
Let's start on this side of the graph. Let's say we take x equal to 0.
00:33:13.176 --> 00:33:16.749
That's 0 over 0 plus 1, so 0.
00:33:20.865 --> 00:33:27.396
We know that as we get close to the vertical asymptote, the behavior should look like this.
00:33:27.403 --> 00:33:32.639
If we are not sure, we can plot another point say where x is negative 1/2.
00:33:33.818 --> 00:33:38.869
Then as we move this direction, we have that horizontal asymptote,
00:33:39.371 --> 00:33:44.879
so we just leave out that hole.
00:33:45.348 --> 00:33:47.652
Again, you can plot as many points as you need to
00:33:47.660 --> 00:33:49.516
to be sure that you have the shape correct.
00:33:51.046 --> 00:33:55.292
What about over here? Let's say this is negative 1,
00:33:55.305 --> 00:34:02.875
so we'll take negative 2.
00:34:04.983 --> 00:34:12.072
This is the point (-2, 2), and we can plot more points again if we need to
00:34:12.175 --> 00:34:13.884
to be sure of the behavior.
00:34:14.389 --> 00:34:18.488
We know that we have a vertical asymptote here going up
00:34:18.814 --> 00:34:20.894
and then the horizontal asymptote here.
00:34:23.030 --> 00:34:25.861
We have a graph of a rational function with a hole in it.
00:34:26.364 --> 00:34:37.618
In calculus, this hole is called a removable discontinuity.
00:34:38.325 --> 00:34:43.515
This means that it's one point that divides the graph into separate pieces,
00:34:44.026 --> 00:34:47.135
but it can be filled in with just that one point
00:34:47.262 --> 00:34:50.340
to make it continuous and smooth on that portion.
00:34:54.785 --> 00:34:56.920
With sketching the graph of a rational function,
00:34:56.940 --> 00:34:59.299
we want to use every tool at our disposal.
00:34:59.575 --> 00:35:01.803
The first thing we want to find is its domain.
00:35:05.862 --> 00:35:15.302
Factoring the denominator, I see that x equals negative 2 or x equals positive 2
00:35:15.348 --> 00:35:18.055
both will make the denominator equal to 0,
00:35:18.214 --> 00:35:24.944
so I'm going to exclude those values from the domain.
00:35:25.395 --> 00:35:29.587
The domain will be all real numbers less than negative 2,
00:35:31.030 --> 00:35:37.859
between negative 2 and positive 2, and greater than 2.
00:35:40.589 --> 00:35:44.990
Next, I want to check for any common factors. There are none.
00:35:45.414 --> 00:36:00.134
This tells me that I have no removable discontinuity.
00:36:01.702 --> 00:36:03.140
Now we'll check for symmetry.
00:36:10.783 --> 00:36:14.867
To check for symmetry, we want to find f of negative x.
00:36:17.609 --> 00:36:25.993
Replacing the x with negative x, I have 1 over negative x squared minus 4.
00:36:25.996 --> 00:36:30.350
Now be very careful with those parentheses. The negative x is what is squared.
00:36:31.937 --> 00:36:37.815
Simplifying, this is the same as 1 over x squared minus 4.
00:36:38.592 --> 00:36:43.874
We see the f of negative x is in fact the same thing as f of x.
00:36:45.855 --> 00:36:54.052
This means that our function is even.
00:36:55.317 --> 00:36:57.304
Now I want to find any intercepts.
00:37:02.000 --> 00:37:05.798
To find the y-intercept, we let x equal 0,
00:37:06.384 --> 00:37:08.558
so that means we want to find f of 0.
00:37:16.097 --> 00:37:19.071
We have a y-intercept at negative 1/4.
00:37:20.988 --> 00:37:31.691
To find the x-intercepts, we want to solve f of x equals 0.
00:37:36.475 --> 00:37:41.216
Multiplying both sides by x squared minus 4, we get 1 equals 0,
00:37:41.308 --> 00:37:52.366
but of course 1 doesn't equal 0. That means that there are no x-intercepts.
00:37:53.977 --> 00:37:56.011
Now we need to find any asymptotes.
00:37:57.946 --> 00:38:03.352
The function will have a vertical asymptote at any value that makes the denominator 0
00:38:03.411 --> 00:38:16.141
but doesn't make the numerator 0, so at negative 2 and at positive 2.
00:38:18.998 --> 00:38:21.573
The horizontal asymptote depends on
00:38:23.008 --> 00:38:27.068
the degree of the numerator related to the degree of the denominator.
00:38:27.531 --> 00:38:32.914
The degree of the denominator m is greater than the degree of the numerator, which is 0,
00:38:34.224 --> 00:38:41.039
so the horizontal asymptote is y equals 0.
00:38:42.488 --> 00:38:45.682
To draw the graph, we'll start with the things we've found so far.
00:38:47.170 --> 00:38:57.162
The vertical asymptotes are x equals 2 and x equals negative 2.
00:39:02.897 --> 00:39:08.187
The horizontal asymptote is y equals 0, which is already there. That's the x-axis.
00:39:09.514 --> 00:39:15.059
Do I have any intercepts? I had one. I have a y-intercept at negative 1/4.
00:39:17.938 --> 00:39:21.178
That's all I know so far, so I'll have to use some test values
00:39:21.186 --> 00:39:23.553
to help me determine the shape of the graph.
00:39:27.054 --> 00:39:30.503
Let's start on the far left side.
00:39:31.647 --> 00:39:37.533
What is the graph doing when I'm close to the vertical asymptote on the left side?
00:39:38.026 --> 00:39:41.282
We'll start with negative 3.
00:39:45.689 --> 00:39:52.039
Substituting that into our function, we have 1 over negative 3 squared minus 4.
00:39:54.738 --> 00:39:58.837
9 minus 4 is 5, so that's 1/5,
00:40:01.852 --> 00:40:04.320
and we need to know if we need to plot more points.
00:40:04.331 --> 00:40:08.229
The answer is: probably, but you can also reason a little bit,
00:40:08.232 --> 00:40:11.086
knowing a little bit about the general shape of your rational functions.
00:40:11.917 --> 00:40:16.929
We can't go across the x-axis because we don't have any more x-intercepts.
00:40:17.267 --> 00:40:19.278
We know that there's supposed to be an asymptote,
00:40:19.289 --> 00:40:26.273
so as x goes towards negative infinity, we're getting close to the asymptote.
00:40:26.557 --> 00:40:30.261
Here, if we maybe plotted one more point in the middle,
00:40:30.269 --> 00:40:32.365
we would see that these values are rising
00:40:32.372 --> 00:40:35.968
and getting to the vertical asymptote from the left.
00:40:38.909 --> 00:40:43.859
Here, we'll need to plot a couple of points. We'll go with negative 1.
00:40:52.704 --> 00:40:56.703
1 minus 4. That would be negative 3, so this is negative 1/3.
00:41:00.469 --> 00:41:03.137
Using the asymptote that I know exists here,
00:41:03.149 --> 00:41:06.490
if I want to plot a few more points and be certain I have the right shape, that's fine.
00:41:07.704 --> 00:41:10.519
Then you might be thinking: I've got more points to plot,
00:41:10.532 --> 00:41:14.623
but, in reality, I have one other piece of information available.
00:41:15.933 --> 00:41:21.571
This is an even function. That means that f of negative x is the same as f of x,
00:41:21.906 --> 00:41:32.975
so I know that if f of negative 3 is 1/5, then f of 3 is also 1/5.
00:41:33.284 --> 00:41:46.906
If f of negative 1 is negative 1/3, this is going to be the same for f of 1.
00:41:48.063 --> 00:41:58.371
Again, I'm using the asymptotes to help. I can plot more points as I need to.
00:41:59.885 --> 00:42:04.417
Okay, it's an even function. It should be symmetric with respect to the y-axis,
00:42:04.425 --> 00:42:06.204
and it certainly looks like it is.
00:42:10.603 --> 00:42:12.591
Let's graph another rational function.
00:42:12.985 --> 00:42:17.452
This is the function f of x equals 3x over x squared minus 25.
00:42:18.683 --> 00:42:28.169
As always, we'll start by finding the domain. Factor the denominator,
00:42:29.350 --> 00:42:33.905
and the numbers that make our denominator equal to 0 will be negative 5 and positive 5.
00:42:34.043 --> 00:42:37.052
These are the numbers to be excluded from the domain of the function.
00:42:41.121 --> 00:42:44.980
Our domain is all real numbers less than negative 5,
00:42:46.384 --> 00:42:52.402
between negative 5 and positive 5, and greater than 5.
00:42:53.675 --> 00:42:57.090
Now look at the numerator. We can't factor, so we see that
00:42:57.113 --> 00:43:00.516
we don't have any common factors between the numerator and the denominator.
00:43:01.276 --> 00:43:15.315
That means we don't have to worry about any removable discontinuities.
00:43:17.956 --> 00:43:28.238
Now let's talk about symmetry. We need to find f of negative x.
00:43:41.423 --> 00:43:46.110
That's 3 times negative x over negative x squared minus 25.
00:43:46.672 --> 00:43:54.254
The numerator is negative 3x. The denominator is the same as x squared minus 25.
00:43:54.948 --> 00:43:59.327
We see that in this example, this is not the original function we had.
00:43:59.599 --> 00:44:02.809
In fact, it's the opposite of the original function we had,
00:44:03.253 --> 00:44:08.466
so f of negative x is the same as negative f of x.
00:44:09.667 --> 00:44:32.393
This means that f is an odd function, and it's symmetric with respect to the origin.
00:44:35.304 --> 00:44:40.462
Now let's look for any intercepts. To find the y-intercept, we let x equal 0.
00:44:40.558 --> 00:44:50.236
We need to find f of 0.
00:44:51.421 --> 00:44:55.922
Substituting 0 for each x and then simplifying, we get 0.
00:44:58.716 --> 00:45:00.781
For the x-intercepts, we've already seen that
00:45:00.789 --> 00:45:03.581
there are no common factors between the numerator and the denominator,
00:45:03.864 --> 00:45:08.945
so the x-intercepts will be any values of x that make our numerator equal to 0.
00:45:13.269 --> 00:45:17.855
We need to solve 0 equals 3x, our numerator,
00:45:18.789 --> 00:45:24.141
so we see that x equals 0 is our x-intercept. Now let's talk about asymptotes.
00:45:30.172 --> 00:45:35.091
Vertical asymptotes occur at any value of x that makes our denominator 0
00:45:35.158 --> 00:45:37.517
that does not also make the numerator 0.
00:45:37.807 --> 00:45:40.583
That's negative 5 and positive 5.
00:45:40.851 --> 00:45:48.358
Our vertical asymptotes are x equals negative 5 and x equals 5.
00:45:49.123 --> 00:45:53.027
Horizontal asymptote: whether it exists and what it is depends on
00:45:53.035 --> 00:45:56.809
the relationship between the degrees of the numerator and the denominator.
00:46:00.602 --> 00:46:05.420
The numerator is degree 1. The denominator is degree 2.
00:46:05.616 --> 00:46:11.448
Since the degree of the denominator m is greater than the degree of the numerator,
00:46:12.504 --> 00:46:15.822
then y equals 0 is your horizontal asymptote.
00:46:20.273 --> 00:46:22.585
Let's put what we know so far on the graph.
00:46:24.018 --> 00:46:42.465
We found vertical asymptotes: x equals 5 and negative 5.
00:46:44.284 --> 00:46:49.342
We found the horizontal asymptote is y equals 0, which is the x-axis.
00:46:51.028 --> 00:46:54.168
The x- and y-intercept are both 0.
00:46:56.550 --> 00:46:59.760
This is all we have so far. We'll have to use test values
00:47:00.624 --> 00:47:03.014
to determine the rest of the shape of the graph.
00:47:04.126 --> 00:47:11.312
Let's start on the right-hand side. Choosing a number bigger than 5, perhaps 6.
00:47:17.494 --> 00:47:20.174
We're going to substitute in our given function.
00:47:31.229 --> 00:47:39.086
11 times 1 is 11, so when x is 6, y is 18/11.
00:47:39.089 --> 00:47:44.015
That's close to 2 but a little bit less than 2.
00:47:47.703 --> 00:47:52.135
Okay, using the asymptotes as a guide and plotting more points if you wish,
00:47:52.579 --> 00:47:55.827
we will graph this section of the function.
00:48:00.035 --> 00:48:03.338
Alright, we found earlier that our function is an odd function,
00:48:03.353 --> 00:48:07.940
and that means that f of negative x is the same as negative f of x.
00:48:09.132 --> 00:48:15.420
Using that, we can find f of negative 6. It has to be the opposite of f of 6.
00:48:22.176 --> 00:48:30.389
We can plot that point, and again using the asymptotes and perhaps plotting more points,
00:48:30.390 --> 00:48:33.637
we can finish the shape of our graph in that section.
00:48:35.786 --> 00:48:37.936
Now what about in between the asymptotes?
00:48:38.965 --> 00:48:47.612
I'll choose x equals 2. Substitute into the given function.
00:48:58.380 --> 00:49:11.916
7 times negative 3 is negative 21, and reducing that is negative 2/7.
00:49:11.994 --> 00:49:16.509
6 divided by 3 is 2.
00:49:17.339 --> 00:49:22.250
21 divided by 3 is 7. Okay, so this is negative 2/7.
00:49:25.846 --> 00:49:29.202
Then finally, using the fact that our function is odd one more time,
00:49:29.927 --> 00:49:46.864
we know that f of negative 2 is negative f of 2, so we get positive 2/7.
00:49:49.031 --> 00:49:55.107
Okay, we know that we have a vertical asymptote here.
00:49:55.613 --> 00:49:59.411
We cross through our x-intercept, and we have another vertical asymptote.
00:49:59.657 --> 00:50:01.931
Again, we can plot as many points as necessary
00:50:01.931 --> 00:50:04.166
to be confident that that is the shape of our graph.
00:50:07.730 --> 00:50:15.161
Let's look at another rational function: f of x equals 3x plus 1 over x squared plus x minus 20.
00:50:15.931 --> 00:50:19.110
We start by finding the domain, so we need to factor the denominator.
00:50:29.276 --> 00:50:37.722
5 times 4 is 20, so if we use plus 5 and minus 4 on the other one, this should work.
00:50:38.558 --> 00:50:42.541
The numbers excluded from the domain will be negative 5 and positive 4.
00:50:48.776 --> 00:50:55.326
Our domain is all real numbers less than negative 5,
00:50:56.641 --> 00:51:03.737
between negative 5 and positive 4, and greater than 4.
00:51:04.981 --> 00:51:07.425
There are no common factors in the numerator and denominator,
00:51:07.432 --> 00:51:09.791
so we don't have any removable discontinuities.
00:51:10.006 --> 00:51:11.607
We'll move on to finding symmetry.
00:51:13.822 --> 00:51:17.642
To find the symmetry, if there is any, we find f of negative x.
00:51:25.396 --> 00:51:37.623
Okay, so replacing x with negative x
00:51:38.374 --> 00:51:49.567
and then cleaning it up a little bit, this is negative 3x plus 1 over x squared minus x minus 20.
00:51:50.186 --> 00:51:53.865
What we notice is that this is not equal to the original function,
00:51:54.321 --> 00:51:57.446
nor is it the opposite of the original function.
00:51:58.100 --> 00:52:06.279
This graph has no symmetry. It's neither even nor odd.
00:52:07.941 --> 00:52:09.597
Now let's talk about intercepts.
00:52:09.985 --> 00:52:17.910
We'll go ahead and find the y-intercept first.
00:52:19.960 --> 00:52:35.620
Let x equal 0, so we're finding f of 0: negative 1/20.
00:52:36.095 --> 00:52:37.712
Now we'll find the x-intercept.
00:52:38.974 --> 00:52:42.586
We saw that there were no common factors in the numerator and the denominator,
00:52:43.148 --> 00:52:46.102
so that means we can find the x-intercepts by finding
00:52:46.628 --> 00:52:56.440
the numbers that make our numerator equal to 0. Solve 3x plus 1 equals 0.
00:52:57.382 --> 00:53:03.132
Our x-intercept is negative 1/3. Now let's find the asymptotes.
00:53:06.382 --> 00:53:16.968
We have vertical asymptotes at x equals negative 5 and x equals 4.
00:53:18.518 --> 00:53:20.885
To find the horizontal asymptote, we look at
00:53:20.892 --> 00:53:25.476
the degree of the numerator and the degree of the denominator and compare them.
00:53:26.683 --> 00:53:30.596
The degree of the denominator is 2. The degree of the numerator is 1.
00:53:31.109 --> 00:53:34.844
Whenever the degree of the denominator m is greater than n,
00:53:35.399 --> 00:53:38.369
then y equals 0 is our horizontal asymptote.
00:53:44.937 --> 00:53:48.626
As you can see, I've gone ahead and sketched the vertical asymptotes:
00:53:48.641 --> 00:53:54.245
x equals negative 5 and x equals 4. Our horizontal asymptote is y equals 0.
00:53:55.100 --> 00:53:59.361
Let's plot our intercepts. The y-intercept is negative 1/20.
00:54:00.971 --> 00:54:06.562
The x-intercept is negative 1/3, and that's all I have.
00:54:06.862 --> 00:54:14.854
I need test values. I've picked four. 5 is in the far right-hand section.
00:54:15.603 --> 00:54:20.267
Substituting 5 into our given function, we come up with the value of 8/5.
00:54:24.551 --> 00:54:29.099
Again, using the asymptotes to guide us as well as plotting as many points as we'd like,
00:54:29.460 --> 00:54:31.223
we can draw that section of the graph.
00:54:33.298 --> 00:54:38.449
x equals 1 is in this section. Substituting that into the given function
00:54:38.761 --> 00:54:41.507
and evaluating it, we come up with negative 2/9.
00:54:47.258 --> 00:54:51.079
Let's go ahead and plot this point when x is negative 1.
00:54:51.344 --> 00:54:55.420
By substituting in the given function, we find the y-value is 1/10.
00:54:56.816 --> 00:54:58.950
Now, what does the graph look like in this portion?
00:54:58.953 --> 00:55:01.158
Well, there's a vertical asymptote here,
00:55:02.051 --> 00:55:05.957
so it seems reasonable that the graph is going to make this sort of shape.
00:55:06.324 --> 00:55:13.107
If we're not convinced of that, we simply need to plot more points here or on this side.
00:55:16.317 --> 00:55:20.494
The last test point we'll consider is x equals negative 6.
00:55:21.328 --> 00:55:26.834
Substituting that into the given function and evaluating this, we come up with negative 17/10.
00:55:30.960 --> 00:55:35.585
Again using the asymptotes to guide us as well as plotting more points as we wish,
00:55:36.236 --> 00:55:37.566
here's that section of the graph.
00:55:38.388 --> 00:55:41.474
You will want to continue to practice these skills over and over
00:55:41.482 --> 00:55:45.589
until you're comfortable doing these examples and others like them comfortably.