### 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.
- 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.
- 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.
- 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.
- 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.
- 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
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.
- 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
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 screen-captured 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 second-degree 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 discontinuity*0239

*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 work--it 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 that--it 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 discontinuity--a 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 right--as it is 2.01--right 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 it--x-values 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 y-values 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 here--let'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 side--it 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: 2.1--y is 21--it is large; 2.01--a little closer to x--y becomes 201; even closer at 2.001--a 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 asymptotes*1232

*tell us what is happening at very large values of x and very, very small values of x--*1241

*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 calculator--that 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 idea*1375

*that when x is 100, it is pretty close; then I get up to x is 1000; y is .5005--it gets closer;*1379

*10,000--y approaches this line; 100,000--it 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; 10,000--barely 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 negative--large 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 domain--extremely 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: 5.99--y 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 6--*2153

*I made x 6.01--y 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 well--that it never quite reaches 0.*2291

*If you plotted additional points of x that were very, very small--very negative--you 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 asymptote--that, 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 happening--that 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 -3.001--very 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 way--what 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 negative--has 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 = 1--horizontal 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 = 5--these are excluded.*2958

*However, since (x + 3) is a common binomial factor, I have a hole here: x = -3--there 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 like--what 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 5--values 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 4.999--now 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 approaching*3390

*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 undefined--*3414

*the function is undefined--at 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, 2016

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 2x-1/x+3.

You then continue to calculate points based on 2x-1/x-5.

You seemed to have made a typo.