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INSTRUCTORSCarleen EatonGrant Fraser
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Lecture Comments (7)

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 2x-1/x+3.

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

You seemed to have made a typo.

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

Describe any holes and vertical asymptotes.
f(x) = [(2x − 6)/(x2 − 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)/(x2 − 2x − 3)] = [(2( − ))/(( − )( + ))]
  • f(x) = [(2x − 6)/(x2 − 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:
Holes: x = 3 Vertical Asymptote: x = − 1
Describe any holes and vertical asymptotes.
f(x) = [( − 3x2 − 3x)/(x2 − 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) = [( − 3x2 − 3x)/(x2 − 1)] = [( − 3x( − ))/(( + )( − ))]
  • f(x) = [( − 3x2 − 3x)/(x2 − 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:
Holes: x = − 1 Vertical Asymptote: x = 1
Graph
f(x) = [(x2 + 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) = [(x2 + x − 6)/( − 4x + 12)] = [(( + )( − ))/( − 4( − ))]
  • f(x) = [(x2 + 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=[(x2+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.
Graph
f(x) = [(x2 − 7x + 12)/(x2 − 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) = [(x2 − 7x + 12)/(x2 − 4)] = [(( − )( − ))/(( + )( − ))]
  • f(x) = [(x2 − 7x + 12)/(x2 − 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.
Graph
f(x) = [(x2 + 3x − 4)/(x2 + 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) = [(x2 + 3x − 4)/(x2 + 2x − 3)] = [(( + )( − ))/(( + )( − ))]
  • f(x) = [(x2 + 3x − 4)/(x2 + 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.
Graph
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.
Graph
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.
Graph
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.
Graph
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.
Graph
f(x) = [2/(x2 − 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/(x2 − 4)] = [2/(( + )( − ))]
  • f(x) = [2/(x2 − 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/(x2−4)]
    -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

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) = (x2 + 8x + 15)/(x2 - 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 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) = (x2 + 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 (x2 - 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 asymptotes1232

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 idea1375

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 2x2, 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 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 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