Graphing Rational Functions
 If the numerator and denominator have a common binomial factor, the graph will have a hole at the point where this factor is 0. If the denominator has a binomial factor that is not in the numerator, the graph will have a vertical asymptote at the point where this factor is 0.
 The graph of a rational function often has a horizontal asymptote.
Graphing Rational Functions
f(x) = [(2x − 6)/(x^{2} − 2x − 3)]
 Holes occur when you have a common binomial factor that cancels out.
 Vertical Asymptotes occur at restricted values in the domain. These are values of x that make the denominator equal to zero.
 Step 1  Factor
 f(x) = [(2x − 6)/(x^{2} − 2x − 3)] = [(2( − ))/(( − )( + ))]
 f(x) = [(2x − 6)/(x^{2} − 2x − 3)] = [(2(x − 3))/((x − 3)(x + 1))]
 Step 2  Identify holes by identifying common binomial factors. Then eliminate the binomial factors.
 The only binomial factor is (x − 3), therefore, there will be a hole at x = 3
 f(x) = [(2(x − 3))/((x − 3)(x + 1))] = [2/((x + 1))]
 f(x) = [2/(x + 1)]
 Step 3  Identify vertical asymptotes by setting denominator equal to zero
 x + 1 = 0
 x = − 1
 Holes:
 Vertical Asymptote:
f(x) = [( − 3x^{2} − 3x)/(x^{2} − 1)]
 Holes occur when you have a common binomial factor that cancels out.
 Vertical Asymptotes occur at restricted values in the domain. These are values of x that make the denominator equal to zero.
 Step 1  Factor
 f(x) = [( − 3x^{2} − 3x)/(x^{2} − 1)] = [( − 3x( − ))/(( + )( − ))]
 f(x) = [( − 3x^{2} − 3x)/(x^{2} − 1)] = [( − 3x(x + 1))/((x + 1)(x − 1))]
 Step 2  Identify holes by identifying common binomial factors. Then eliminate the binomial factors.
 The only binomial factor is (x + 1), therefore, there will be a hole at x = − 1
 f(x) = [( − 3x)/((x − 1))]
 f(x) = [( − 3x)/(x − 1)]
 Step 3  Identify vertical asymptotes by setting denominator equal to zero
 x − 1 = 0
 x = 1
 Holes:
 Vertical Asymptote:
f(x) = [(x^{2} + x − 6)/( − 4x + 12)]
 Holes occur when you have a common binomial factor that cancels out.
 Vertical Asymptotes occur at restricted values in the domain. These are values of x that make the denominator equal to zero.
 Step 1  Factor
 f(x) = [(x^{2} + x − 6)/( − 4x + 12)] = [(( + )( − ))/( − 4( − ))]
 f(x) = [(x^{2} + x − 6)/( − 4x + 12)] = [((x + 3)(x − 2))/( − 4(x − 3))]
 Step 2  Identify holes by identifying common binomial factors. Then eliminate the binomial factors.
 There are no holes in this graph.
 Step 3  Identify vertical asymptotes by setting denominator equal to zero
 x − 3 = 0
 x = 3
 Step 4  Graph the vertical asymptotes and create a table of values, with values of x to theleft and right of the vertical asymptote

x y=[(x^{2}+x−6)/(−4x+12)] 5 0.44 3 0 0 0.5 1 0.5 2 0 2.75 4.31 3.25 7.81 4 3.5 5 3 6 3 7 3.13 8 3.3  Step 5  Sketch the graph. Play close attention as you move closer to the vertical asymptote.
 The graph should never cross the vertical asymptote.
f(x) = [(x^{2} − 7x + 12)/(x^{2} − 4)]
 Holes occur when you have a common binomial factor that cancels out.
 Vertical Asymptotes occur at restricted values in the domain. These are values of x that make
 the denominator equal to zero.
 Step 1  Factor
 f(x) = [(x^{2} − 7x + 12)/(x^{2} − 4)] = [(( − )( − ))/(( + )( − ))]
 f(x) = [(x^{2} − 7x + 12)/(x^{2} − 4)] = [((x − 3 )(x − 4))/((x + 2)(x − 2))]
 Step 2  Identify holes by identifying common binomial factors. Then eliminate the binomial factors.
 There are no holes in this graph.
 Step 3  Identify vertical asymptotes by setting denominator equal to zero
 x + 2 = 0 and x − 2 = 0
 x = − 2 and x = 2
 Step 4  Graph the vertical asymptotes and create a table of values, with values of x to theleft and right of the vertical asymptotes

x y=[((x−3)(x−4))/((x+2)(x−2))] 6 2.81 4 4.67 3 8.4 1.5 14.14 0 3 1 2 1.5 2.14 1.95 10.9 2.05 9.15 3 0 4 0 6 0.19  Step 5  Sketch the graph. Play close attention as you move closer to the vertical asymptote.
 The graph should never cross the vertical asymptote.
f(x) = [(x^{2} + 3x − 4)/(x^{2} + 2x − 3)]
 Holes occur when you have a common binomial factor that cancels out.
 Vertical Asymptotes occur at restricted values in the domain. These are values of x that make the denominator equal to zero.
 Step 1  Factor
 f(x) = [(x^{2} + 3x − 4)/(x^{2} + 2x − 3)] = [(( + )( − ))/(( + )( − ))]
 f(x) = [(x^{2} + 3x − 4)/(x^{2} + 2x − 3)] = [((x + 4)(x − 1))/((x + 3)(x − 1))]
 Step 2  Identify holes by identifying common binomial factors. Then eliminate the binomial factors.
 Notice how there's only one hole, x = 1
 f(x) = [((x + 4)(x − 1))/((x + 3)(x − 1))] = [((x + 4))/((x + 3))]
 Step 3  Identify vertical asymptotes by setting denominator equal to zero
 x + 3 = 0 x = 3
 Step 4  Sketch the graph. Play close attention as you move closer to the vertical asymptote.
 The graph should never cross the vertical asymptote.
f(x) = [x/( − 3x + 6)]
 Holes occur when you have a common binomial factor that cancels out.
 Vertical Asymptotes occur at restricted values in the domain. These are values of x that make the denominator equal to zero.
 Step 1  Factor
 f(x) = [x/( − 3x + 6)] = [x/( − 3( − ))]
 f(x) = [x/( − 3x + 6)] = [x/( − 3(x − 2))]
 Step 2  Identify holes by identifying common binomial factors. Then eliminate the binomial factors.
 Notice how there are no holes.
 Step 3  Identify vertical asymptotes by setting denominator equal to zero
 x − 2 = 0
 x = 2
 Step 4  Graph the vertical asymptotes and create a table of values, with values of x to theleft and right of the vertical asymptotes

x [x/(−3x+6)] 6 0.25 4 0.22 2 0.17 0 0 1 0.33 1.75 2.33 2.25 3 2.75 1.22 4 0.67 6 0.5 8 0.44 10 0.42  Step 5  Sketch the graph. Play close attention as you move closer to the vertical asymptote.
 The graph should never cross the vertical asymptote.
f(x) = [x/(4x − 8)]
 Holes occur when you have a common binomial factor that cancels out.
 Vertical Asymptotes occur at restricted values in the domain. These are values of x that make the denominator equal to zero.
 Step 1  Factor
 f(x) = [x/(4x − 8)] = [x/(4( − ))]
 f(x) = [x/(4x − 8)] = [x/(4(x − 2))]
 Step 2  Identify holes by identifying common binomial factors. Then eliminate the binomial factors.
 Notice how there are no holes.
 Step 3  Identify vertical asymptotes by setting denominator equal to zero
 x − 2 = 0
 x = 2
 Step 4  Graph the vertical asymptotes and create a table of values, with values of x to the left and right of the vertical asymptotes

x [x/(4x+8)] 6 0.19 4 0.17 2 0.13 0 0 1 0.25 1.75 1.75 2.25 2.25 2.75 0.92 4 0.5 6 0.38 8 0.33 10 0.31  Step 5  Sketch the graph. Play close attention as you move closer to the vertical asymptote.
 The graph should never cross the vertical asymptote.
f(x) = [3/(x − 2)]
 Holes occur when you have a common binomial factor that cancels out.
 Vertical Asymptotes occur at restricted values in the domain. These are values of x that make the denominator equal to zero.
 Step 1  Factor
 In this case, there is nothing to factor out.
 Step 2  Identify holes by identifying common binomial factors. Then eliminate the binomial factors.
 Notice how there are no holes.
 Step 3  Identify vertical asymptotes by setting denominator equal to zero
 x − 2 = 0
 x = 2
 Step 4  Graph the vertical asymptotes and create a table of values, with values of x to theleft and right of the vertical asymptotes

x [3/(x−2)] 6 0.38 4 0.05 2 0.75 0 1.5 1 3 1.75 12 2.25 12 2.75 4 4 1.4 6 0.75 8 0.50 10 0.38  Step 5  Sketch the graph. Play close attention as you move closer to the vertical asymptote.
 The graph should never cross the vertical asymptote.
f(x) = [( − 3x − 6)/(x + 1)]
 Holes occur when you have a common binomial factor that cancels out.
 Vertical Asymptotes occur at restricted values in the domain. These are values of x that make the denominator equal to zero.
 Step 1  Factor
 f(x) = [( − 3x − 6)/(x + 1)] = [( − 3(x + ))/(x + 1)]
 f(x) = [( − 3x − 6)/(x + 1)] = [( − 3(x + 2))/(x + 1)]
 Step 2  Identify holes by identifying common binomial factors. Then eliminate the binomial factors.
 Notice how there are no holes.
 Step 3  Identify vertical asymptotes by setting denominator equal to zero
 x + 1 = 0
 x = − 1
 Step 4  Graph the vertical asymptotes and create a table of values, with values of x to theleft and right of the vertical asymptotes

x y=[(−3x−6)/(x+1)] 8 2.57 4 2 2 0 1.5 3 1.25 9 0.75 15 0.5 9 0 6 2 4 4 3.6 6 3.43 8 3.33  Step 5  Sketch the graph. Play close attention as you move closer to the vertical asymptote.
 The graph should never cross the vertical asymptote.
f(x) = [2/(x^{2} − 4)]
 Holes occur when you have a common binomial factor that cancels out.
 Vertical Asymptotes occur at restricted values in the domain. These are values of x that make the denominator equal to zero.
 Step 1  Factor
 f(x) = [2/(x^{2} − 4)] = [2/(( + )( − ))]
 f(x) = [2/(x^{2} − 4)] = [2/((x + 2)(x − 2))]
 Step 2  Identify holes by identifying common binomial factors. Then eliminate the binomial factors.
 Notice how there are no holes.
 Step 3  Identify vertical asymptotes by setting denominator equal to zero
 x + 2 = 0 and x − 2 = 0
 x = − 2 and x = 2
 Step 4  Graph the vertical asymptotes and create a table of values, with values of x to theleft and right of the vertical asymptotes

x y=[2/(x^{2}−4)] 6 0.06 4 0.17 3 0.4 2.25 1.88 2.1 4.88 1.9 5.13 0 0.5 1.9 5.13 2.1 4.88 3 0.4 4 0.17 6 0.06  Step 5  Sketch the graph. Play close attention as you move closer to the vertical asymptote.
 The graph should never cross the vertical asymptote.
*These practice questions are only helpful when you work on them offline on a piece of paper and then use the solution steps function to check your answer.
Answer
Graphing Rational Functions
Lecture Slides are screencaptured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.
 Intro 0:00
 Rational Functions 0:18
 Restriction
 Example: Rational Function
 Breaks in Continuity 2:52
 Example: Continuous Function
 Discontinuities
 Example: Excluded Values
 Graphs and Discontinuities 5:02
 Common Binomial Factor (Hole)
 Example: Common Factor
 Asymptote
 Example: Vertical Asymptote
 Horizontal Asymptotes 20:00
 Example: Horizontal Asymptote
 Example 1: Holes and Vertical Asymptotes 26:12
 Example 2: Graph Rational Faction 28:35
 Example 3: Graph Rational Faction 39:23
 Example 4: Graph Rational Faction 47:28
Algebra 2
Transcription: Graphing Rational Functions
Welcome to Educator.com.0000
We have been working with rational expressions; and now we are going to talk about graphing rational functions.0002
And graphs of rational functions have some features that you may not have seen so far, when you were working with other types of graphs.0008
OK, so first, defining a rational function: a rational function is in the form f(x) = p(x)/q(x),0018
where both p(x) and q(x) are polynomial functions, and q(x) is not equal to 0.0026
As usual, we have the restriction that the denominator cannot be 0.0034
Values that make the function in the denominator equal to 0 are excluded from the domain of this function.0040
For example, f(x) = (x^{2} + 8x + 15)/(x^{2}  2x  8):0050
we have worked with rational expressions so far, and now we are just talking about these as functions.0063
Working with a function like this, I would look at the denominator and factor that out.0070
This gives me x...and since there is a negative here, it is plus and minus.0079
Factors of 8 are 1 and 8, and 2 and 4; and I am looking for a set of factors that is going to add up to 2.0086
And that would be 2  4; so the 4 goes in the negative place, and the 2 in the positive.0097
So, I have factored this out; and the reason is to look for the excluded values, values that would make q(x) negative.0106
I am going to use the zero product property, because if this entire expression is 0,0116
that could occur if either x + 2 = 0 or x  4 = 0, or they are both equal to 0.0123
Using the zero product property, I have x + 2 = 0 and x  4 = 0.0130
This becomes x = 2; and this right here is x = 4; therefore, my excluded values are x = 2 and x = 4.0139
So, excluded from the domain of f(x) are x = 2 and x = 4.0151
OK, so now, working with these rational functions, we are just going to do some basic graphing, first introducing the concept of breaks and continuity.0166
We have worked with linear functions; we have worked with polynomial functions, particularly quadratic functions,0176
which are a seconddegree polynomial function; and we see that these graphs are always continuous.0183
For example, I may have had a graph of a polynomial like this, or of a quadratic function like this.0190
So, there are no areas of the graph where there are missing pieces, or it stops and then starts again: these graphs are always continuous.0201
Rational functions are different: they may have points at which they are not continuous, and these are called discontinuities.0211
And there are a couple of different types of discontinuities.0219
These discontinuities occur because there are excluded values from the domain.0222
When we worked with, say, a quadratic function, we may have said, "The domain is all real numbers."0228
There are no parts of the graph where the function is not defined.0235
Now, we are going to see a couple of different situations; we are going to see one type of discontinuity0239
that we will discuss in a minute, that is called a hole, where the graph is going along,0247
and then suddenly it is not defined in this certain section.0255
We are also going to see a couple of other, more complicated, types of discontinuities, called asymptotes,0259
where the graph will approach a certain value, but it will not quite reach it.0266
And we will define all that in a second; but there are a couple of types of discontinuities.0271
These would occur at excluded values: for example, f(x) = (x^{2} + 3x  1)/(x + 8).0275
Well, x = 8 is an excluded value; therefore, there will be a discontinuity here.0286
And we are going to talk right now about the different types and how to know which type of discontinuity you are dealing with.0295
OK, there are two ways that a graph of a rational function can show discontinuity.0302
If the function in the numerator, p(x), and the function in the denominator, q(x), have a common binomial factor,0308
then the graph has a hole at the point of the discontinuity.0316
Remember that we defined f(x) as consisting of p(x)/q(x).0319
So, if these two have a common factor, then we will see a hole at that point in the graph.0325
For example, let's let f(x) equal (x^{2}  9)/(2x  6).0331
We are going to factor both the numerator and the denominator, because I want to find the excluded values,0342
but I also want to see if there is a common factor.0349
So, this, as usual, factors into (x + 3) (x  3); in the denominator, there is a common factor of 2, so it factors into (x  3).0354
Looking here, I have (x  3) here and (x  3) here, so let's just focus on that for right now.0368
And I have a common binomial factor, (x  3); I know that, if (x  3) equals 0, this will equal 0.0378
And therefore, when x  3 equals 0, this will be undefined.0402
Well, when x = 3, then I would have 3 here; 3  3 is 0; that doesn't workit is undefined.0410
Therefore, x = 3 is an excluded value, and there is going to be a discontinuity here.0417
There will be a discontinuity at x = 3; and the type of discontinuity is a hole (at x = 3).0425
And the reason it is a hole is because there is a common factor; this is a common binomial factor.0434
If there was an excluded value here, but it wasn't a common factor, we will get a different discontinuity that we will talk about in a second.0440
So, let's just go ahead and find some points and see what this graph is going to look like.0447
Now, I factored this out; I left my factors here, so I could look at it and see that, yes, this is a common factor.0450
But for the purposes of graphing, I am not going to leave this factor here.0459
I want to make my graphing as simple as possible, so let me rewrite that up here: f(x) = (x + 3)(x  3)/2(x  3).0463
I have figured out that there is a hole at x = 3; now, I am done with this factor.0476
I am just going to cancel it out so that I can make my graphing and my calculations much simpler,0479
because this is just going to be f(x) = (x + 3)/2.0486
Therefore, as usual when I graph, I can find some points.0491
Let's start out with 5, because 5 + 3 is 2, divided by 2 is 1.0498
3 + 3 is 0, divided by 2 is 0; 1 + 3 is 2, divided by 2 is 1.0508
Remember that 3 looks fine here, but back in the original I saw that that would make the denominator 0.0520
So, I have to remember that 3 is an excluded value.0528
I can't use this for the domain; I can't find a function value for thatit is undefined.0535
When x is 5, 5 plus 3 is 8, divided by 2 is 4; this is enough for me to go ahead and graph.0543
So, when x is 5, y is 1; when x is 3, y is 0; when x is 1, y is 1; when x is 3, I have an excluded value; I can't do anything with that.0549
I will show you how I will represent it in a second.0567
When x is 5, y is 4.0569
This gives me enough to draw a line; however, when I get to 3, I am just going to leave a circle there, indicating that there is a hole at x = 3.0573
This graph is discontinuous; and there is a discontinuity here at x = 3, since that value is excluded from the domain.0593
OK, so this is our first type of discontinuitya hole.0602
The second type is called an asymptote.0607
Recall that we said a rational function would be something like f(x) = p(x)/q(x).0614
If there is a common factor between these two, then we get a hole.0624
Now, there may be excluded values that don't involve a common binomial factor.0628
For example, if there is a factor in the denominator ax  b, that the numerator does not have,0634
there, at that point, we will have a vertical asymptote wherever the point is that we solve for x.0643
We took this ax  b and went ahead and solved for x to find this excluded value.0651
That is where we are going to have a vertical asymptote.0657
An asymptote is a line that the graph approaches, but it never crosses.0660
And these can be either horizontal lines or vertical; and we are focusing on vertical lines right now.0664
For example, let f(x) equal x/(x  2).0669
Here is a factor in the denominator, (x  2), a binomial factor that is not present in the numerator.0675
So, it is not a hole; instead, there is going to be a vertical asymptote.0681
My excluded value is going to be x = 2, because x  2 cannot equal 0.0686
If it does, this would be undefined; so x = 2 is excluded from the domain, and there is a vertical asymptote here.0694
When you find the asymptote, the first thing you should do is actually go ahead and put a vertical line at that point,0711
because otherwise you risk a situation where you might accidentally cross it.0722
So, I know that there is a vertical asymptote at x = 2, so I am just going to go ahead and draw a dotted line here,0730
which is going to tell me that my graph can approach this line, but it cannot cross it.0737
In order to know what is going on, what you want to do is look at points on either side of 2,0744
and figure out what the graph does as x approaches 2, as it is very close to 2 from the left,0751
as it is 1.999, and as it is very close from the rightas it is 2.01right around here, in addition to other points.0759
Let's just start out with some points, say, over here at 2.0768
When x is 2, then we are going to get...2 and 2 is 4, so 2/4 is going to give me positive 1/2.0782
OK, so when x is 1, that is going to be 1/3; that is going to be positive 1/3.0794
When x is 0, that is going to be 0 divided by (0  2); well, 0 divided by anything is 0.0803
When x is 1, 1 divided by 1  2 is going to be 1 divided by 1, so that is going to be 1.0811
So, I know that x is never going to equal 2; so I can't just say, "OK, the next point I am going to find is x = 2."0836
I am going to find points close to itxvalues very close to it, but not actually equal to 2.0846
So, let's start from this left side and think about what happens when, say, x is 1.9.0855
Well, when x is 1.9, then that is going to give me 1.9  2, so that is going to give me .1.0866
So, if I just do 1.9 divided by .1 and move the decimals over, that is going to give me 19; I'll put that right here.0877
OK, so now, let's think about x getting even closer to 2: let's let x be 1.99.0887
So, it is approaching 2 from this left side: that is going to give me 1.99; 1.99  2 is going to be .01; so that is going to give me 199.0896
And you can continue on, and you will see the pattern that, as x approaches 2 from the left side, the yvalues become very large negative numbers.0916
OK, so I am doing a little bit of graphing: let's get some points and just think about what this is going to look like, coming at it from this side.0929
Let's graph out some points: when x is 2, y is 1/2; when x is 1, y is 1/3; when x is 0, y is 0; 1, 1.0943
OK, now when x gets close here, this is going to be way down here, so I can't represent it exactly.0957
But what I do know is that, the closer this x is getting, the more and more large of a negative number y is.0966
So, I have the general shape here that...what is happening is: this is approaching the asymptote, but it is not going to cross it.0974
I see that there is a discontinuity right here, because x can never quite reach 2; and so, the graph is not ever going to reach this line.0991
OK, now, this is part of the graph: this is what happens when x approaches 2 from the left.1008
I also want to figure out what happens when x approaches 2 from the right.1015
So, let's look at some values that are just a little bit greater than 2, such as 2.1.1020
OK, so when x is 2.1, 2.1  2 is going to be .1, so that is going to be 21.1028
Now, getting a little bit closer: 2.1, say, is right herelet's get a little bit closer: let's make this 2.01.1040
That is going to give me 2.01 divided by .01, or 201.1050
OK, getting even closer, let's make x 2.001; so I am coming at this from this sideit is 2.001, divided by .001, which is going to give me 2001.1056
And then (that is right in here), I am going to graph some more points out here,1073
just so I have more of the shape of the graph, beyond just this little area.1076
OK, let's let x equal 4; 4 divided by 4  2 (4 divided by 2) is 2, and also 3: 3 divided by 3  2 (that is 3 divided by 1) is 3.1082
First, out here, I have 3, 4; so when x is 4, y is 2; when x is 3, y is 3.1098
Now, the closer I get to this...at 2.1, y is going to be way up here at 21; as I get even closer, y is going to be even bigger.1114
So, what is happening with this graph is: as the values of x approach 2, y becomes very large.1123
And this graph is going to approach this line, but it is not going to cross it.1136
So, a couple of things that we notice: one is that, as x approaches 2 from the left, the values for y become very large in the negative direction.1145
I am closer and closer and closer to 2, but never reaching it.1157
I jump to the other side of 2: 2.1y is 21it is large; 2.01a little closer to xy becomes 201; even closer at 2.001a very large value for y.1161
So, this is what the graph is going to look like, because there is a vertical asymptote right here at x = 2.1176
And what you will see is that the graph is going to approach from this side and not quite reach that value;1184
and it is going to approach from this side and not quite reach the value.1189
So, there is a discontinuity at the vertical line x = 2.1192
There are also horizontal asymptotes: the graph of a rational function can have both vertical asymptotes and horizontal asymptotes.1201
And these asymptotes occur at values that are excluded from the range of f(x).1213
So, this is going to be a line; that is horizontal line defined by a value y = something.1218
So, let's look at this function: g(x) = (x + 1)/2x.1225
I am just going to focus on the horizontal asymptote; and the important thing is that horizontal asymptotes1232
tell us what is happening at very large values of x and very, very small values of x1241
large positive values, or values way over here that are very negative.1247
So, I am not going to worry about the middle of the graph right now; I am just going to focus on what is happening at the extremes of the domain.1251
Let's let x equal, first, a very large number: 100.1259
That is going to give me 101/200; and if you were to figure that out (you may end up using your calculatorthat is fine) that will give you .505.1266
OK, so let's make x even bigger; let's make it 1000, because I am trying to figure out what is happening at the extreme right side of the graph.1277
This is going to give me...this was x = 100; so if x = 1000, I am going to get 1001/2000, and if you figure that out, it is going to come out to .5005.1286
When x is 10,000, you are going to end up with .50005; if x is 100,000, it is .500005.1309
What you can see happening is that, as x becomes very large, y is approaching .5, but it is never quite getting there.1328
What that tells me is that there is a horizontal asymptote at y = .5.1338
So, I am just going to go ahead and call this .5, and put my horizontal asymptote right there, and do a sketch of the rest of the graph.1348
This is going to be .5, and there is a line here at y = .5.1357
And what I see happening is that...let's make this 100, and jump up to 1,000, and then 10,000, and then 100,000;1364
of course, if this were proportional, it would be much longer, but this gives you the general idea1375
that when x is 100, it is pretty close; then I get up to x is 1000; y is .5005it gets closer;1379
10,000y approaches this line; 100,000it is approaching it even closer.1392
So, what is happening (actually, it is approaching it from above): when x is 100, we are going to be slightly above .5, at .505.1398
When x is 1000, now it is at .5005, just a little bit above .5; 10,000barely above it; 100,000; and so on.1412
So, at very large values of x, the graph approaches y = .5, but it doesn't quite reach it.1425
Therefore, y = .5 is a horizontal asymptote.1437
Let's look at what is happening at values of x that are very negativelarge negative values of x.1447
Let's look at 100: and again, you may need to use a calculator to calculate this out,1459
to give you the idea: x = 100; therefore, this is going to give me 99/200.1465
Calculating that out, it comes out to .495.1475
When x equals 1000, this is going to give me 999, divided by 2000.1479
And then, I divide that; I am going to get a positive number, .4995.1487
And continuing on my calculations with 10,000, this would give me .49995, and so on.1495
So, you can see what happens: as x is very negative, y approaches .5, but it doesn't quite reach it.1504
So, if I made this 100, 1000, 10,000, and so on, I see that, at 100, this is pretty close;1512
it is y = .495; but then I get to a bigger number, like 1000; it is even closer at .4995.1523
A bigger number is even closer; so I can see that, on this side, as x becomes very negative, y is approaching .5 from below.1532
So here, at very large values of x, the graph approaches .5 from above; at very small values of x, the graph approaches .5 from below.1543
It is the same idea as a vertical asymptote, with the approaching, but never crossing that line.1553
Horizontal asymptotes tell you what is happening at the extremes of the domainextremely large values and extremely small values.1559
This first example just asks us to describe any holes and vertical asymptotes that this function would have.1574
We don't need to actually graph it out.1580
So remember: to find those, you are going to have to factor and look at excluded values.1582
The denominator is already factored for us.1588
All right, in the numerator, I have a negative here, so plus and minus.1597
I have factors of 12: 1 and 12, 2 and 6, 3 and 4; and I see that I have +1: I need these factors to add up to 1.1602
So, I am going to look for factors close together: 3 and 4.1616
Since this is positive, I am going to make the 4 positive: 4  3 is going to give me 1.1620
So, I am going to factor this out to (x + 4) (x  3).1625
Now, recall that a hole will occur when you have a common binomial factor.1631
And I do have that: (x + 4) is a common binomial factor.1637
So, I am going to go ahead and look at the excluded values that I have: using the zero product property, I have (x + 4) (x  7) = 0.1640
And this would be a situation that is not allowed: so I need to find out what values of x would create this situation.1652
x + 4 = 0 and x  7 = 0: either of those will cause this whole thing to become 0.1658
So, x = 4 and x = 7 are excluded values; they are excluded from the domain.1666
Now, since (x + 4) is a common binomial factor in the numerator and denominator, there is going to be a hole at x = 4.1676
x = 7 is also an excluded value; however, there is not a common factor.1688
So, it is simply going to be a vertical asymptote at x = 7.1701
There is a hole, where there is a common factor, for the value of x that would create a zero down here, which is x = 4,1713
and a vertical asymptote at the other excluded value of x = 7.1721
And if you were to graph that, you would start out by finding these, so that you would be aware of that for your graph.1726
So now, we are asked to graph this rational function; and as always, I am going to start out by looking for holes and vertical asymptotes.1737
I am going to factor the denominator; and this gives me...since I have a negative here, I have (x + something) (x  something).1750
The only integer factors of 5 are 1 and 5, and I know that if I add + 1 and negative...1759
Actually, we are looking at 6, so it is 1 and 6, 2 and 3.1769
And I am looking for factors that will add up to 5; and I look at 6 and 1.1774
If I took 1  6, that is 5; so this is going to factor out to (x + 1) (x  6).1780
(x + 1) times (x  6) is not allowed to equal 0; if it does, then this is undefined.1793
So, excluded values are going to be values of x that cause this product to be 0.1799
Using the zero product property, I get these two equations, and I find that x = 1 and x = 6 are excluded from the domain.1804
Since x + 1 is a common factor in the numerator and denominator, there is going to be a hole at x = 1.1818
And there is going to be a vertical asymptote at x = 6.1828
Put in the vertical asymptote represented by a dashed line, so that I know that my graph will approach, but it will not cross.1841
And then, I also have to remember that wherever the graph ends up, x = 1 is going to be excluded.1851
So, I will put an open circle there to denote that.1859
I am just starting out with some values, and especially focusing on values as x gets very close to 6,1865
either from the left (5.99999) or from the right (6.0001), thinking about what happens right around this asymptote.1873
Let's just start out with some values that aren't quite that close; but say x is 2.1885
Now, in order to make my life simpler, I am actually going to, now that I have looked at these factors...1893
I don't need that factor anymore, so I can cancel this out.1899
And I am just going to end up with 1/(x  6), because remember, this really means 1 times (x + 1).1902
So, I am canceling this out, and I get 1/(x  6)much easier to work with for the graph.1910
When x is 2, that is going to give me 1/(1/8); when x is 1, I can't forget that this is not defined.1916
I don't have a value for when x is 1; this function is just not defined.1928
When x is 0, that is going to give me 1/6; so that is 1/6.1937
Now, let's think about what happens as we get closer to 6: let's let x be 5.1946
That is 1/(5  6), so that is 1/1; that is 1.1953
I am starting to graph a few of these points: when x is 2, y is 1/8, just right down here.1960
When x is 1, y is not defined; when x is 0, this is going to be 1/6.1976
When x is 5, y is going to be 1; so the general shape is going to be like this, so far.1986
But I am going to make sure, here at 1, that I show that there is a hole at x = 1.1994
Now, I want to figure out what is going on right here, so I am going to pick some values even closer to 6.2007
I am going to pick 5.9: well, if x is 5.9, that is going to give me 1/(5.9  6); that is going to be 1/.1, so that is going to give me 10.2012
I want to get even closer to 6, so let's try 5.9.2034
Now, when I have x = 5.99, this is going to give me 1/(5.99  6), which is .01; so this is going to give me 100.2047
So, what you can see is happening is that, as x approaches 6 (this is actually supposed to be 1)...here we had y as 1 when x is 5.2068
Then, x is 5.9, which is closer to 6; x becomes 2, 4, 6, 8, 10; so it is way down here.2086
Now, even closer to 6: 5.99y becomes a very large negative value, 100.2097
So, I can see that what is happening is that this graph is going to approach this line, but it is never going to reach it; it is never going to cross it.2105
Something else I also need to look at is what is happening to the graph as x approaches 6 from the other side.2128
Let's look at values just on the other side of 6: let's look at 6.1.2137
And if you work this out, you will see that, when x is 6.1, y is 10.2145
When x is 6.01 (when x is 6.1, I am going to see that x is 2, 4, 6, 8, 10; y is way up here)if I got even closer to 62153
I made x 6.01y is going to become 100.2170
So, I already see this usual trend of approaching, but not crossing, the vertical asymptote.2175
So, as x approaches 6 from the right side, y becomes very large.2183
As x approaches 6 from the left side, y becomes very, very large negative numbers.2190
To get a better sense of the graph, let's also look at some numbers over here.2197
7 would be right about there; so when x is 7, that is 1/(7  6), so that would be 1; when x is 7, y is 1.2201
When x is, let's say, 10, that is going to give me 1/(10  6); that is 1/4.2212
When x is 7, y is going to be 1; when x is 10, it is going to be right here; it is just going to be like this.2220
So, the other thing that you notice as you plot more points: let's plot a very large value of x.2232
Let's plot 100: so this is 1/(100  6), so that is 1/96, and that is going to give me .01.2246
Something else that you are noticing here on the graph, as you plot more points, is that,2255
in addition to this vertical asymptote, there is also a horizontal asymptote at y = 0.2260
And you can see that, as x gets larger, the graph approaches 0, but it doesn't quite reach 0.2279
And if you plotted additional large points, you would see that, as wellthat it never quite reaches 0.2291
If you plotted additional points of x that were very, very smallvery negativeyou would see the same thing:2298
that the graph is going to approach values of the function that equal 0, but it is never going to quite get there.2303
Reviewing what we found about this graph: we found that there is going to be a hole at x = 1, designated by an open circle.2313
There is a vertical asymptote at x = 6; that is an excluded value, x = 6.2321
So, x is never going to equal 6; and as x approaches 6, y becomes very, very large or very, very small, as it gets near that graph.2332
But it is never going to cross the graph: x will never equal 6, so the graph will never cross that line.2344
We also see that there is a horizontal asymptotethat, for very large negative or large positive values of x,2349
the graph is going to approach this line, but it is never going to cross it.2356
OK, in this example, we are going to be graphing f(x) = x/(x + 3).2362
So, the first thing is to figure out excluded values: x + 3 cannot equal 0, so I am going to look for what value of x would cause this to become 0.2369
And that would be x = 3; and that is excluded.2383
And since there is not a common binomial factor, there is going to be a vertical asymptote here.2386
So, I am just going to start out by marking that, so we don't lose track of that.2394
There are no holes, because there are no common binomial factors.2400
Just to get a sense of the graph, I am going to start out by plotting some points here, approaching x = 3.2404
And then, I am going to go to the other side and do the same thing.2414
Let's start over on this side, with values such as 2.2420
When x is 2, if you figure this out, it comes out to say that y is also 2.2424
When x is 1, we are going to get y equaling 1/2.2432
When x is 1, 1 over 4 would give me 1/4; when x is 0, then I am going to get 0 divided by something, which is 0.2440
Now, of course, I am not going to use 3 as a value, because that is an excluded value.2453
Let's go ahead and plot these: this is 2 and 2; 1 and 1/2; 0 and 0; and 1 and 1/4.2461
So, I can already see what this graph is approaching here.2478
What happens when x gets very close to 3, a little bit to the right of it (values like 2.9)?2481
It is coming at it from this way: 2.9; 2.99; 2.999; what happens there?2493
Well, when x is 2.9, if you figure this out, you will see x = 2.9; that is going to give 2.9/(2.9 + 3); that is going to be .1; that is going to give me 29.2499
So, as x approaches 3, y becomes very large in the negative direction.2522
And just to verify that, taking another point, 2.99 is going to give me 2.99/.01, equals 299.2529
So, that is enough to give me the trend of what is happeningthat this graph is curving like this.2545
And at values very close to 3, a little bit larger than 3 (a little bit less negative), y becomes very large in the negative direction.2553
OK, now I am going to jump over to the other side of this asymptote and figure out what is going on at values over here on the left:2569
5, 4, and then things like 3.01 or 3.001very close on this left side.2576
Let's make a separate column for that and start out with something such as 5.2585
When x equals 5, that is going to give me 5/(5 + 3), so that is 2, so that is 5/2, or 2 and 1/2.2596
At 4, that is going to give me 4/(4 + 3), so 4/1; 4/1 is just going to be 4.2612
That is over here; now, that is 5; 5/2 is right here; 4 is 4; OK.2626
Now, what is going to happen, then, since the graph is moving up this waywhat is going to happen very close to 3?2641
I can already predict that y is likely going to get very large, and approach, but not cross, the graph.2650
But let's go ahead and verify that.2656
Values close to 3, but just left of it, would be something like 3.1.2659
And if you work that out, you will see that that comes out to 31.2667
If you pick a value that is even closer to 3, like 3.01, you will get 301.2672
If I go even closer, 3.001, I will get 3001.2685
So, over here, as the graph approaches from the left, y becomes very, very negativehas very large negative values.2691
As the graph approaches x = 3 from the right, the values of x become very large.2700
Now, the one other thing we want to look at is, "Are there any horizontal asymptotes?"2710
And I can notice here that my graph is sort of flattening out when I get near 1.2716
So, I have some values over here, 2, 1...but I want to look at much larger values of x, just to see what is happening.2722
For example, now I am looking for what is going on at the extreme values of x, when the domain is large, or for domain values that are very small.2737
When x is 100, what I am going to get is 100 divided by 103, which is .97.2747
Let's make x even bigger: so, if x is 1000, I am going to get 1000 divided by 1003, which is .997.2755
And you can already see that, as x gets very large, y is approaching 1; but it is not actually reaching it.2764
What this is telling me is that I have a horizontal asymptote right here: y = 1horizontal asymptote.2774
To verify that: what I expect to happen is that, on this side, the graph is also going to approach y = 1, but it is not going to cross it.2788
So, let's look at what happens at very negative values of x, just picking a value like 1000.2797
So, that is going to give me 1000/(997); and calculating that out, you would get 1.003.2805
So again, I see that, at very large negative values of x, this graph is coming in close to 1, but it is not quite reaching it.2818
OK, so to sum up: this graph has two branches to it, and it has a vertical asymptote at x = 3; it has a horizontal asymptote at y = 1.2829
We see the graph approaching both of those asymptotes, but not crossing them.2842
This function starts out looking pretty complicated; so let's factor it and see what we have.2849
All right, so my leading coefficient is 2, so I have a 2 right here and an x here.2858
I have a negative here, so I am going to have one negative and one positive; but I need to figure out which way is correct.2863
Factors of 3 are just 1 and 3; and I have 2x and x; let's try some combinations.2870
If I put 2x  1 here and x + 3 there, what do I get?2880
Well, I get 2x^{2}, and then I get 6x  x  3; 6x and x is 5x, so this is the correct factorization, (2x  1) (x + 3).2886
In the denominator, I have a leading coefficient of 1, so that makes it easier; and I have a negative here: +, .2908
Factors of 15 are 1 and 15, 3 and 5; and I want factors that are going to add up to 2.2918
So, 3  5 equals 2, so this is the correct factorization.2928
What you will see, then, is excluded values: (x + 3) (x  5)I cannot allow this to equal zero.2937
So, using the zero product property, I know that, if x + 3 equals 0, or x  5 = 0, then this whole thing will end up being zero.2946
Solving for x: there are excluded values at x = 3 and x = 5these are excluded.2958
However, since (x + 3) is a common binomial factor, I have a hole here: x = 3there is a hole right here.2967
x = 5: since (x  5) is not a common factor of the numerator, then what I have here is a vertical asymptote at x = 5.2980
2, 4, 6...x = 5 is going to be here; let's draw in our vertical line.2995
The graph is going to approach this line from either side, but it is not going to cross it, because x can never equal 5.3010
And I can't forget that, at x = 3, there is going to be a hole; but I don't know where the graph is going to cross yet, so I can't draw that in.3018
OK, I am going to start out finding some values, just to get a sense of the graph.3028
And then, I am going to hone in on this region.3033
When x is 8, let's figure out why: to make my graphing easier, I am going to cancel out common factors,3036
because now I have done what I need to do with those factors.3046
I am going to cancel out that common factor of (x + 3); and what I am going to graph is (2x  1)/(x + 3).3049
Now, when x is 8, if you plugged that in and figured it out, you would find that y is 5.3057
So, when x is 8, y is 5; getting closer to 5, but not quite reaching it, let's try 4.9.3066
Again, if you were to plug that in and do the math, the calculations, you would find that y then becomes...3082
actually, just to the right...let's do slightly different values.3095
We are looking just to the right of 5, so we are going to look at values that are larger than 5, values such as 5.1.3098
So, we will work on the left in a minute.3106
We have this point here; now we are looking just greater than 5, at things like 5.1.3107
If we let x equal 5.1, we would find that y equals 92.3114
So, as I get very close to x, y is going to get very large.3120
Let's try honing in even closer: when x is something like 5.01, then y gets even larger at 902.3126
And if I continued on, I would find that this gets even larger and larger, as I add 5.001...I am going to get even larger values.3136
x is almost 5, and then y becomes a very large value.3147
So, that is what it looks like, right there.3152
Now, way out to the right, what does it look likewhat is the value of the function when x is very large?3157
Looking at something like when x is 100, calculating that out, you would find that y is 2.09.3167
OK, then let's make x even larger: x is 1000: let's figure out what y is going to be.3176
Well, y comes out to 2.009; again, you can see what is happening.3187
As x becomes very large, y is approaching 2, but it is not going to get there.3195
So, we know that what we have is a horizontal asymptote at y = 2.3202
because as x gets larger and larger, it is going to approach, but it will not cross.3209
All right, so I covered what is going on over here to the right of this asymptote; I also determined that I have a horizontal asymptote at y = 2.3222
And let's go over to the left, to values less than x = 5, and find out what is happening.3230
First, I am just picking values right on this side of 5values such as 4.9.3237
Plug in 4.9 here, and you will find that y is 88.3249
So, when x is just slightly less than 5, y is going to be a very negative value down here.3254
As x gets even closer to 5, letting x be something like 4.99, y is going to become even larger, but in that negative direction.3263
Or 4.999now we are very close here to x = 5, and y is getting very large; but it is not quite reaching.3276
And that is what we expected.3290
Now, I can also just take some other points to get a more accurate graph.3294
Let's let x equal 0; if x equals 0, then I am going to get 0  1 (that is 1), over 0 + 3; so this is going to be 1/3.3302
Now, I have a better sense of where to draw this line.3314
And what I also know is that this part of the graph is going to approach this horizontal asymptote, but it is not going to reach it.3325
And I could verify that by finding some very negative values of x, such as 100.3339
And when x is 100, y is .98.3345
So, I can verify that, the bigger I make x...it is going to approach this asymptote, but it is never going to quite reach it.3356
So again, the pertinent points are a vertical asymptote at x = 5, and something else we talked about:3365
at 3, there is a hole, so I have to draw that in; so here x is 3, so I can't have a value here.3373
I have to just draw a circle, because the graph is actually undefined right there.3382
Let's see, there is a vertical asymptote here, a horizontal asymptote here, and a graph approaching3390
both the horizontal and vertical asymptotes, but never actually reaching it.3409
And the same over here on this side; and make sure that you denote that the graph is undefined3414
the function is undefinedat the excluded value of x = 3.3423
That concludes this section of Educator.com on graphing rational functions.3428
1 answer
Last reply by: Khanh Nguyen
Sun Jan 3, 2016 6:37 PM
Post by Khanh Nguyen on January 3 at 06:34:54 PM
At 32:48, for x = 5, 5  6 = 1. So the chart should have added a  sign to the 1.
You forgot to add the  sign.
2 answers
Last reply by: Kavita Agrawal
Wed Jun 19, 2013 10:07 PM
Post by Christine Kuhlman on September 17, 2012
Is there a time that a line could cross a horizontal asymptote? If so, how can you find out when and by how much it does.
I thought that with asymptotes the line could never touch it, but today my teacher told us they can cross and I'm very confused.
1 answer
Last reply by: Dr Carleen Eaton
Sat Feb 26, 2011 6:02 PM
Post by Edgar Rariton on February 26, 2011
Regarding example IV, you say that you're going to cancel out the common factors of x+3, but then you say to graph 2x1/x+3.
You then continue to calculate points based on 2x1/x5.
You seemed to have made a typo.