WEBVTT mathematics/algebra-2/eaton
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Welcome to Educator.com.
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We have been working with rational expressions; and now we are going to talk about graphing rational functions.
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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.
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OK, so first, defining a rational function: a **rational function** is in the form f(x) = p(x)/q(x),
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where both p(x) and q(x) are polynomial functions, and q(x) is not equal to 0.
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As usual, we have the restriction that the denominator cannot be 0.
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Values that make the function in the denominator equal to 0 are excluded from the domain of this function.
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For example, f(x) = (x² + 8x + 15)/(x² - 2x - 8):
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we have worked with rational expressions so far, and now we are just talking about these as functions.
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Working with a function like this, I would look at the denominator and factor that out.
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This gives me x...and since there is a negative here, it is plus and minus.
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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.
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And that would be 2 - 4; so the 4 goes in the negative place, and the 2 in the positive.
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So, I have factored this out; and the reason is to look for the excluded values, values that would make q(x) negative.
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I am going to use the zero product property, because if this entire expression is 0,
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that could occur if either x + 2 = 0 or x - 4 = 0, or they are both equal to 0.
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Using the zero product property, I have x + 2 = 0 and x - 4 = 0.
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This becomes x = -2; and this right here is x = 4; therefore, my excluded values are x = -2 and x = 4.
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So, excluded from the domain of f(x) are x = -2 and x = 4.
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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.
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We have worked with linear functions; we have worked with polynomial functions, particularly quadratic functions,
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which are a second-degree polynomial function; and we see that these graphs are always continuous.
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For example, I may have had a graph of a polynomial like this, or of a quadratic function like this.
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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.
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Rational functions are different: they may have points at which they are not continuous, and these are called discontinuities.
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And there are a couple of different types of discontinuities.
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These discontinuities occur because there are excluded values from the domain.
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When we worked with, say, a quadratic function, we may have said, "The domain is all real numbers."
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There are no parts of the graph where the function is not defined.
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Now, we are going to see a couple of different situations; we are going to see one type of discontinuity
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that we will discuss in a minute, that is called a **hole**, where the graph is going along,
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and then suddenly it is not defined in this certain section.
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We are also going to see a couple of other, more complicated, types of discontinuities, called **asymptotes**,
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where the graph will approach a certain value, but it will not quite reach it.
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And we will define all that in a second; but there are a couple of types of discontinuities.
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These would occur at excluded values: for example, f(x) = (x² + 3x - 1)/(x + 8).
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Well, x = -8 is an excluded value; therefore, there will be a discontinuity here.
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And we are going to talk right now about the different types and how to know which type of discontinuity you are dealing with.
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OK, there are two ways that a graph of a rational function can show discontinuity.
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If the function in the numerator, p(x), and the function in the denominator, q(x), have a common binomial factor,
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then the graph has a hole at the point of the discontinuity.
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Remember that we defined f(x) as consisting of p(x)/q(x).
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So, if these two have a common factor, then we will see a hole at that point in the graph.
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For example, let's let f(x) equal (x² - 9)/(2x - 6).
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We are going to factor both the numerator and the denominator, because I want to find the excluded values,
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but I also want to see if there is a common factor.
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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).
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Looking here, I have (x - 3) here and (x - 3) here, so let's just focus on that for right now.
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And I have a common binomial factor, (x - 3); I know that, if (x - 3) equals 0, this will equal 0.
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And therefore, when x - 3 equals 0, this will be undefined.
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Well, when x = 3, then I would have 3 here; 3 - 3 is 0; that doesn't work--it is undefined.
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Therefore, x = 3 is an excluded value, and there is going to be a discontinuity here.
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There will be a discontinuity at x = 3; and the type of discontinuity is a hole (at x = 3).
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And the reason it is a hole is because there is a common factor; this is a common binomial factor.
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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.
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So, let's just go ahead and find some points and see what this graph is going to look like.
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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.
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But for the purposes of graphing, I am not going to leave this factor here.
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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).
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I have figured out that there is a hole at x = 3; now, I am done with this factor.
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I am just going to cancel it out so that I can make my graphing and my calculations much simpler,
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because this is just going to be f(x) = (x + 3)/2.
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Therefore, as usual when I graph, I can find some points.
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Let's start out with -5, because -5 + 3 is -2, divided by 2 is -1.
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-3 + 3 is 0, divided by 2 is 0; -1 + 3 is 2, divided by 2 is 1.
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Remember that 3 looks fine here, but back in the original I saw that that would make the denominator 0.
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So, I have to remember that 3 is an excluded value.
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I can't use this for the domain; I can't find a function value for that--it is undefined.
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When x is 5, 5 plus 3 is 8, divided by 2 is 4; this is enough for me to go ahead and graph.
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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.
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I will show you how I will represent it in a second.
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When x is 5, y is 4.
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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.
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This graph is discontinuous; and there is a discontinuity here at x = 3, since that value is excluded from the domain.
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OK, so this is our first type of discontinuity--a hole.
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The second type is called an asymptote.
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Recall that we said a rational function would be something like f(x) = p(x)/q(x).
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If there is a common factor between these two, then we get a hole.
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Now, there may be excluded values that don't involve a common binomial factor.
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For example, if there is a factor in the denominator ax - b, that the numerator does not have,
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there, at that point, we will have a vertical asymptote wherever the point is that we solve for x.
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We took this ax - b and went ahead and solved for x to find this excluded value.
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That is where we are going to have a vertical asymptote.
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An asymptote is a line that the graph approaches, but it never crosses.
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And these can be either horizontal lines or vertical; and we are focusing on vertical lines right now.
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For example, let f(x) equal x/(x - 2).
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Here is a factor in the denominator, (x - 2), a binomial factor that is not present in the numerator.
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So, it is not a hole; instead, there is going to be a vertical asymptote.
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My excluded value is going to be x = 2, because x - 2 cannot equal 0.
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If it does, this would be undefined; so x = 2 is excluded from the domain, and there is a vertical asymptote here.
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When you find the asymptote, the first thing you should do is actually go ahead and put a vertical line at that point,
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because otherwise you risk a situation where you might accidentally cross it.
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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,
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which is going to tell me that my graph can approach this line, but it cannot cross it.
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In order to know what is going on, what you want to do is look at points on either side of 2,
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and figure out what the graph does as x approaches 2, as it is very close to 2 from the left,
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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.
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Let's just start out with some points, say, over here at -2.
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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.
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OK, so when x is -1, that is going to be -1/-3; that is going to be positive 1/3.
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When x is 0, that is going to be 0 divided by (0 - 2); well, 0 divided by anything is 0.
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When x is 1, 1 divided by 1 - 2 is going to be 1 divided by -1, so that is going to be -1.
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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."
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I am going to find points close to it--x-values very close to it, but not actually equal to 2.
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So, let's start from this left side and think about what happens when, say, x is 1.9.
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Well, when x is 1.9, then that is going to give me 1.9 - 2, so that is going to give me -.1.
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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.
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OK, so now, let's think about x getting even closer to 2: let's let x be 1.99.
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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.
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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.
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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.
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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.
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OK, now when x gets close here, this is going to be way down here, so I can't represent it exactly.
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But what I do know is that, the closer this x is getting, the more and more large of a negative number y is.
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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.
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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.
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OK, now, this is part of the graph: this is what happens when x approaches 2 from the left.
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I also want to figure out what happens when x approaches 2 from the right.
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So, let's look at some values that are just a little bit greater than 2, such as 2.1.
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OK, so when x is 2.1, 2.1 - 2 is going to be .1, so that is going to be 21.
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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.
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That is going to give me 2.01 divided by .01, or 201.
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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.
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And then (that is right in here), I am going to graph some more points out here,
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just so I have more of the shape of the graph, beyond just this little area.
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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.
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First, out here, I have 3, 4; so when x is 4, y is 2; when x is 3, y is 3.
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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.
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So, what is happening with this graph is: as the values of x approach 2, y becomes very large.
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And this graph is going to approach this line, but it is not going to cross it.
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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.
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I am closer and closer and closer to 2, but never reaching it.
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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.
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So, this is what the graph is going to look like, because there is a vertical asymptote right here at x = 2.
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And what you will see is that the graph is going to approach from this side and not quite reach that value;
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and it is going to approach from this side and not quite reach the value.
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So, there is a discontinuity at the vertical line x = 2.
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There are also horizontal asymptotes: the graph of a rational function can have both vertical asymptotes and horizontal asymptotes.
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And these asymptotes occur at values that are excluded from the range of f(x).
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So, this is going to be a line; that is horizontal line defined by a value y = something.
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So, let's look at this function: g(x) = (x + 1)/2x.
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I am just going to focus on the horizontal asymptote; and the important thing is that horizontal asymptotes
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tell us what is happening at very large values of x and very, very small values of x--
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large positive values, or values way over here that are very negative.
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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.
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Let's let x equal, first, a very large number: 100.
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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.
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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.
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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.
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When x is 10,000, you are going to end up with .50005; if x is 100,000, it is .500005.
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What you can see happening is that, as x becomes very large, y is approaching .5, but it is never quite getting there.
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What that tells me is that there is a horizontal asymptote at y = .5.
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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.
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This is going to be .5, and there is a line here at y = .5.
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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;
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of course, if this were proportional, it would be much longer, but this gives you the general idea
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that when x is 100, it is pretty close; then I get up to x is 1000; y is .5005--it gets closer;
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10,000--y approaches this line; 100,000--it is approaching it even closer.
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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.
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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.
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So, at very large values of x, the graph approaches y = .5, but it doesn't quite reach it.
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Therefore, y = .5 is a horizontal asymptote.
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Let's look at what is happening at values of x that are very negative--large negative values of x.
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Let's look at -100: and again, you may need to use a calculator to calculate this out,
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to give you the idea: x = -100; therefore, this is going to give me -99/-200.
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Calculating that out, it comes out to .495.
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When x equals -1000, this is going to give me -999, divided by -2000.
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And then, I divide that; I am going to get a positive number, .4995.
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And continuing on my calculations with -10,000, this would give me .49995, and so on.
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So, you can see what happens: as x is very negative, y approaches .5, but it doesn't quite reach it.
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So, if I made this -100, -1000, -10,000, and so on, I see that, at -100, this is pretty close;
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it is y = .495; but then I get to a bigger number, like -1000; it is even closer at .4995.
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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.
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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.
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It is the same idea as a vertical asymptote, with the approaching, but never crossing that line.
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Horizontal asymptotes tell you what is happening at the extremes of the domain--extremely large values and extremely small values.
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This first example just asks us to describe any holes and vertical asymptotes that this function would have.
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We don't need to actually graph it out.
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So remember: to find those, you are going to have to factor and look at excluded values.
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The denominator is already factored for us.
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All right, in the numerator, I have a negative here, so plus and minus.
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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.
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So, I am going to look for factors close together: 3 and 4.
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Since this is positive, I am going to make the 4 positive: 4 - 3 is going to give me 1.
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So, I am going to factor this out to (x + 4) (x - 3).
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Now, recall that a hole will occur when you have a common binomial factor.
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And I do have that: (x + 4) is a common binomial factor.
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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.
00:27:32.200 --> 00:27:38.500
And this would be a situation that is not allowed: so I need to find out what values of x would create this situation.
00:27:38.500 --> 00:27:46.600
x + 4 = 0 and x - 7 = 0: either of those will cause this whole thing to become 0.
00:27:46.600 --> 00:27:55.800
So, x = -4 and x = 7 are excluded values; they are excluded from the domain.
00:27:55.800 --> 00:28:08.600
Now, since (x + 4) is a common binomial factor in the numerator and denominator, there is going to be a hole at x = -4.
00:28:08.600 --> 00:28:21.100
x = 7 is also an excluded value; however, there is not a common factor.
00:28:21.100 --> 00:28:33.200
So, it is simply going to be a vertical asymptote at x = 7.
00:28:33.200 --> 00:28:40.900
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,
00:28:40.900 --> 00:28:45.700
and a vertical asymptote at the other excluded value of x = 7.
00:28:45.700 --> 00:28:57.400
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.
00:28:57.400 --> 00:29:09.900
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.
00:29:09.900 --> 00:29:19.600
I am going to factor the denominator; and this gives me...since I have a negative here, I have (x + something) (x - something).
00:29:19.600 --> 00:29:29.500
The only integer factors of 5 are 1 and 5, and I know that if I add + 1 and negative...
00:29:29.500 --> 00:29:34.000
Actually, we are looking at 6, so it is 1 and 6, 2 and 3.
00:29:34.000 --> 00:29:40.300
And I am looking for factors that will add up to -5; and I look at 6 and 1.
00:29:40.300 --> 00:29:53.100
If I took 1 - 6, that is -5; so this is going to factor out to (x + 1) (x - 6).
00:29:53.100 --> 00:29:58.900
(x + 1) times (x - 6) is not allowed to equal 0; if it does, then this is undefined.
00:29:58.900 --> 00:30:04.600
So, excluded values are going to be values of x that cause this product to be 0.
00:30:04.600 --> 00:30:18.600
Using the zero product property, I get these two equations, and I find that x = -1 and x = 6 are excluded from the domain.
00:30:18.600 --> 00:30:28.300
Since x + 1 is a common factor in the numerator and denominator, there is going to be a hole at x = 1.
00:30:28.300 --> 00:30:41.300
And there is going to be a vertical asymptote at x = 6.
00:30:41.300 --> 00:30:51.200
Put in the vertical asymptote represented by a dashed line, so that I know that my graph will approach, but it will not cross.
00:30:51.200 --> 00:30:59.000
And then, I also have to remember that wherever the graph ends up, x = -1 is going to be excluded.
00:30:59.000 --> 00:31:05.100
So, I will put an open circle there to denote that.
00:31:05.100 --> 00:31:13.000
I am just starting out with some values, and especially focusing on values as x gets very close to 6,
00:31:13.000 --> 00:31:25.000
either from the left (5.99999) or from the right (6.0001), thinking about what happens right around this asymptote.
00:31:25.000 --> 00:31:33.300
Let's just start out with some values that aren't quite that close; but say x is -2.
00:31:33.300 --> 00:31:38.900
Now, in order to make my life simpler, I am actually going to, now that I have looked at these factors...
00:31:38.900 --> 00:31:42.600
I don't need that factor anymore, so I can cancel this out.
00:31:42.600 --> 00:31:50.600
And I am just going to end up with 1/(x - 6), because remember, this really means 1 times (x + 1).
00:31:50.600 --> 00:31:56.000
So, I am canceling this out, and I get 1/(x - 6)--much easier to work with for the graph.
00:31:56.000 --> 00:32:08.500
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.
00:32:08.500 --> 00:32:17.100
I don't have a value for when x is -1; this function is just not defined.
00:32:17.100 --> 00:32:26.000
When x is 0, that is going to give me 1/-6; so that is -1/6.
00:32:26.000 --> 00:32:32.900
Now, let's think about what happens as we get closer to 6: let's let x be 5.
00:32:32.900 --> 00:32:39.800
That is 1/(5 - 6), so that is 1/1; that is 1.
00:32:39.800 --> 00:32:56.300
I am starting to graph a few of these points: when x is -2, y is -1/8, just right down here.
00:32:56.300 --> 00:33:05.800
When x is -1, y is not defined; when x is 0, this is going to be -1/6.
00:33:05.800 --> 00:33:14.200
When x is 5, y is going to be 1; so the general shape is going to be like this, so far.
00:33:14.200 --> 00:33:26.900
But I am going to make sure, here at -1, that I show that there is a hole at x = -1.
00:33:26.900 --> 00:33:32.100
Now, I want to figure out what is going on right here, so I am going to pick some values even closer to 6.
00:33:32.100 --> 00:33:53.700
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.
00:33:53.700 --> 00:34:07.100
I want to get even closer to 6, so let's try 5.9.
00:34:07.100 --> 00:34:28.100
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.
00:34:28.100 --> 00:34:45.900
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.
00:34:45.900 --> 00:34:56.900
Then, x is 5.9, which is closer to 6; x becomes -2, 4, 6, 8, 10; so it is way down here.
00:34:56.900 --> 00:35:05.500
Now, even closer to 6: 5.99--y becomes a very large negative value, -100.
00:35:05.500 --> 00:35:28.200
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.
00:35:28.200 --> 00:35:36.800
Something else I also need to look at is what is happening to the graph as x approaches 6 from the other side.
00:35:36.800 --> 00:35:44.900
Let's look at values just on the other side of 6: let's look at 6.1.
00:35:44.900 --> 00:35:52.800
And if you work this out, you will see that, when x is 6.1, y is 10.
00:35:52.800 --> 00:36:10.400
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--
00:36:10.400 --> 00:36:14.700
I made x 6.01--y is going to become 100.
00:36:14.700 --> 00:36:23.100
So, I already see this usual trend of approaching, but not crossing, the vertical asymptote.
00:36:23.100 --> 00:36:29.700
So, as x approaches 6 from the right side, y becomes very large.
00:36:29.700 --> 00:36:36.700
As x approaches 6 from the left side, y becomes very, very large negative numbers.
00:36:36.700 --> 00:36:41.200
To get a better sense of the graph, let's also look at some numbers over here.
00:36:41.200 --> 00:36:51.700
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.
00:36:51.700 --> 00:37:00.200
When x is, let's say, 10, that is going to give me 1/(10 - 6); that is 1/4.
00:37:00.200 --> 00:37:11.700
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.
00:37:11.700 --> 00:37:26.300
So, the other thing that you notice as you plot more points: let's plot a very large value of x.
00:37:26.300 --> 00:37:35.400
Let's plot 100: so this is 1/(100 - 6), so that is 1/96, and that is going to give me .01.
00:37:35.400 --> 00:37:40.000
Something else that you are noticing here on the graph, as you plot more points, is that,
00:37:40.000 --> 00:37:58.900
in addition to this vertical asymptote, there is also a horizontal asymptote at y = 0.
00:37:58.900 --> 00:38:11.100
And you can see that, as x gets larger, the graph approaches 0, but it doesn't quite reach 0.
00:38:11.100 --> 00:38:18.600
And if you plotted additional large points, you would see that, as well--that it never quite reaches 0.
00:38:18.600 --> 00:38:23.600
If you plotted additional points of x that were very, very small--very negative--you would see the same thing:
00:38:23.600 --> 00:38:33.600
that the graph is going to approach values of the function that equal 0, but it is never going to quite get there.
00:38:33.600 --> 00:38:41.400
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.
00:38:41.400 --> 00:38:52.100
There is a vertical asymptote at x = 6; that is an excluded value, x = 6.
00:38:52.100 --> 00:39:03.900
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.
00:39:03.900 --> 00:39:09.100
But it is never going to cross the graph: x will never equal 6, so the graph will never cross that line.
00:39:09.100 --> 00:39:16.400
We also see that there is a horizontal asymptote--that, for very large negative or large positive values of x,
00:39:16.400 --> 00:39:22.600
the graph is going to approach this line, but it is never going to cross it.
00:39:22.600 --> 00:39:29.400
OK, in this example, we are going to be graphing f(x) = x/(x + 3).
00:39:29.400 --> 00:39:42.800
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.
00:39:42.800 --> 00:39:45.700
And that would be x = -3; and that is excluded.
00:39:45.700 --> 00:39:54.600
And since there is not a common binomial factor, there is going to be a vertical asymptote here.
00:39:54.600 --> 00:40:00.400
So, I am just going to start out by marking that, so we don't lose track of that.
00:40:00.400 --> 00:40:03.900
There are no holes, because there are no common binomial factors.
00:40:03.900 --> 00:40:13.700
Just to get a sense of the graph, I am going to start out by plotting some points here, approaching x = -3.
00:40:13.700 --> 00:40:19.900
And then, I am going to go to the other side and do the same thing.
00:40:19.900 --> 00:40:24.500
Let's start over on this side, with values such as -2.
00:40:24.500 --> 00:40:32.200
When x is -2, if you figure this out, it comes out to say that y is also -2.
00:40:32.200 --> 00:40:40.200
When x is -1, we are going to get y equaling -1/2.
00:40:40.200 --> 00:40:53.500
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.
00:40:53.500 --> 00:41:01.100
Now, of course, I am not going to use -3 as a value, because that is an excluded value.
00:41:01.100 --> 00:41:17.800
Let's go ahead and plot these: this is -2 and -2; -1 and -1/2; 0 and 0; and 1 and 1/4.
00:41:17.800 --> 00:41:21.500
So, I can already see what this graph is approaching here.
00:41:21.500 --> 00:41:33.000
What happens when x gets very close to -3, a little bit to the right of it (values like -2.9)?
00:41:33.000 --> 00:41:39.600
It is coming at it from this way: -2.9; -2.99; -2.999; what happens there?
00:41:39.600 --> 00:42:02.500
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.
00:42:02.500 --> 00:42:09.000
So, as x approaches -3, y becomes very large in the negative direction.
00:42:09.000 --> 00:42:25.600
And just to verify that, taking another point, -2.99 is going to give me -2.99/.01, equals -299.
00:42:25.600 --> 00:42:33.100
So, that is enough to give me the trend of what is happening--that this graph is curving like this.
00:42:33.100 --> 00:42:49.000
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.
00:42:49.000 --> 00:42:56.300
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:
00:42:56.300 --> 00:43:05.500
-5, -4, and then things like -3.01 or -3.001--very close on this left side.
00:43:05.500 --> 00:43:15.800
Let's make a separate column for that and start out with something such as -5.
00:43:15.800 --> 00:43:31.700
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.
00:43:31.700 --> 00:43:46.600
At -4, that is going to give me -4/(-4 + 3), so -4/-1; -4/-1 is just going to be 4.
00:43:46.600 --> 00:44:01.300
That is over here; now, that is -5; 5/2 is right here; -4 is 4; OK.
00:44:01.300 --> 00:44:10.000
Now, what is going to happen, then, since the graph is moving up this way--what is going to happen very close to -3?
00:44:10.000 --> 00:44:16.500
I can already predict that y is likely going to get very large, and approach, but not cross, the graph.
00:44:16.500 --> 00:44:19.500
But let's go ahead and verify that.
00:44:19.500 --> 00:44:26.800
Values close to -3, but just left of it, would be something like -3.1.
00:44:26.800 --> 00:44:32.500
And if you work that out, you will see that that comes out to 31.
00:44:32.500 --> 00:44:44.700
If you pick a value that is even closer to -3, like -3.01, you will get 301.
00:44:44.700 --> 00:44:51.200
If I go even closer, -3.001, I will get 3001.
00:44:51.200 --> 00:45:00.400
So, over here, as the graph approaches from the left, y becomes very, very negative--has very large negative values.
00:45:00.400 --> 00:45:10.100
As the graph approaches x = -3 from the right, the values of x become very large.
00:45:10.100 --> 00:45:16.100
Now, the one other thing we want to look at is, "Are there any horizontal asymptotes?"
00:45:16.100 --> 00:45:22.400
And I can notice here that my graph is sort of flattening out when I get near 1.
00:45:22.400 --> 00:45:37.200
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.
00:45:37.200 --> 00:45:47.300
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.
00:45:47.300 --> 00:45:55.600
When x is 100, what I am going to get is 100 divided by 103, which is .97.
00:45:55.600 --> 00:46:03.800
Let's make x even bigger: so, if x is 1000, I am going to get 1000 divided by 1003, which is .997.
00:46:03.800 --> 00:46:13.900
And you can already see that, as x gets very large, y is approaching 1; but it is not actually reaching it.
00:46:13.900 --> 00:46:27.900
What this is telling me is that I have a horizontal asymptote right here: y = 1--horizontal asymptote.
00:46:27.900 --> 00:46:36.800
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.
00:46:36.800 --> 00:46:45.100
So, let's look at what happens at very negative values of x, just picking a value like -1000.
00:46:45.100 --> 00:46:57.700
So, that is going to give me -1000/(-997); and calculating that out, you would get 1.003.
00:46:57.700 --> 00:47:08.900
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.
00:47:08.900 --> 00:47:21.700
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.
00:47:21.700 --> 00:47:28.900
We see the graph approaching both of those asymptotes, but not crossing them.
00:47:28.900 --> 00:47:38.000
This function starts out looking pretty complicated; so let's factor it and see what we have.
00:47:38.000 --> 00:47:43.600
All right, so my leading coefficient is 2, so I have a 2 right here and an x here.
00:47:43.600 --> 00:47:49.700
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.
00:47:49.700 --> 00:47:59.700
Factors of 3 are just 1 and 3; and I have 2x and x; let's try some combinations.
00:47:59.700 --> 00:48:05.800
If I put 2x - 1 here and x + 3 there, what do I get?
00:48:05.800 --> 00:48:28.300
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).
00:48:28.300 --> 00:48:38.000
In the denominator, I have a leading coefficient of 1, so that makes it easier; and I have a negative here: +, -.
00:48:38.000 --> 00:48:48.400
Factors of 15 are 1 and 15, 3 and 5; and I want factors that are going to add up to -2.
00:48:48.400 --> 00:48:56.800
So, 3 - 5 equals -2, so this is the correct factorization.
00:48:56.800 --> 00:49:06.200
What you will see, then, is excluded values: (x + 3) (x - 5)--I cannot allow this to equal zero.
00:49:06.200 --> 00:49:18.300
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.
00:49:18.300 --> 00:49:27.500
Solving for x: there are excluded values at x = -3 and x = 5--these are excluded.
00:49:27.500 --> 00:49:39.900
However, since (x + 3) is a common binomial factor, I have a hole here: x = -3--there is a hole right here.
00:49:39.900 --> 00:49:55.400
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.
00:49:55.400 --> 00:50:10.600
2, 4, 6...x = 5 is going to be here; let's draw in our vertical line.
00:50:10.600 --> 00:50:18.300
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.
00:50:18.300 --> 00:50:27.700
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.
00:50:27.700 --> 00:50:33.400
OK, I am going to start out finding some values, just to get a sense of the graph.
00:50:33.400 --> 00:50:36.200
And then, I am going to hone in on this region.
00:50:36.200 --> 00:50:46.600
When x is 8, let's figure out why: to make my graphing easier, I am going to cancel out common factors,
00:50:46.600 --> 00:50:49.100
because now I have done what I need to do with those factors.
00:50:49.100 --> 00:50:56.900
I am going to cancel out that common factor of (x + 3); and what I am going to graph is (2x - 1)/(x + 3).
00:50:56.900 --> 00:51:06.600
Now, when x is 8, if you plugged that in and figured it out, you would find that y is 5.
00:51:06.600 --> 00:51:22.000
So, when x is 8, y is 5; getting closer to 5, but not quite reaching it, let's try 4.9.
00:51:22.000 --> 00:51:35.400
Again, if you were to plug that in and do the math, the calculations, you would find that y then becomes...
00:51:35.400 --> 00:51:38.500
actually, just to the right...let's do slightly different values.
00:51:38.500 --> 00:51:46.100
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.
00:51:46.100 --> 00:51:47.400
So, we will work on the left in a minute.
00:51:47.400 --> 00:51:53.800
We have this point here; now we are looking just greater than 5, at things like 5.1.
00:51:53.800 --> 00:52:00.000
If we let x equal 5.1, we would find that y equals 92.
00:52:00.000 --> 00:52:05.700
So, as I get very close to x, y is going to get very large.
00:52:05.700 --> 00:52:16.100
Let's try honing in even closer: when x is something like 5.01, then y gets even larger at 902.
00:52:16.100 --> 00:52:27.500
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.
00:52:27.500 --> 00:52:32.600
x is almost 5, and then y becomes a very large value.
00:52:32.600 --> 00:52:36.900
So, that is what it looks like, right there.
00:52:36.900 --> 00:52:47.600
Now, way out to the right, what does it look like--what is the value of the function when x is very large?
00:52:47.600 --> 00:52:56.100
Looking at something like when x is 100, calculating that out, you would find that y is 2.09.
00:52:56.100 --> 00:53:07.100
OK, then let's make x even larger: x is 1000: let's figure out what y is going to be.
00:53:07.100 --> 00:53:14.900
Well, y comes out to 2.009; again, you can see what is happening.
00:53:14.900 --> 00:53:22.500
As x becomes very large, y is approaching 2, but it is not going to get there.
00:53:22.500 --> 00:53:29.100
So, we know that what we have is a horizontal asymptote at y = 2.
00:53:29.100 --> 00:53:42.000
because as x gets larger and larger, it is going to approach, but it will not cross.
00:53:42.000 --> 00:53:50.400
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.
00:53:50.400 --> 00:53:57.300
And let's go over to the left, to values less than x = 5, and find out what is happening.
00:53:57.300 --> 00:54:09.400
First, I am just picking values right on this side of 5--values such as 4.9.
00:54:09.400 --> 00:54:13.900
Plug in 4.9 here, and you will find that y is -88.
00:54:13.900 --> 00:54:22.800
So, when x is just slightly less than 5, y is going to be a very negative value down here.
00:54:22.800 --> 00:54:36.400
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.
00:54:36.400 --> 00:54:50.100
Or 4.999--now we are very close here to x = 5, and y is getting very large; but it is not quite reaching.
00:54:50.100 --> 00:54:54.500
And that is what we expected.
00:54:54.500 --> 00:55:01.700
Now, I can also just take some other points to get a more accurate graph.
00:55:01.700 --> 00:55:14.500
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.
00:55:14.500 --> 00:55:25.600
Now, I have a better sense of where to draw this line.
00:55:25.600 --> 00:55:39.000
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.
00:55:39.000 --> 00:55:45.300
And I could verify that by finding some very negative values of x, such as -100.
00:55:45.300 --> 00:55:56.100
And when x is -100, y is .98.
00:55:56.100 --> 00:56:04.700
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.
00:56:04.700 --> 00:56:13.000
So again, the pertinent points are a vertical asymptote at x = 5, and something else we talked about:
00:56:13.000 --> 00:56:22.400
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.
00:56:22.400 --> 00:56:29.700
I have to just draw a circle, because the graph is actually undefined right there.
00:56:29.700 --> 00:56:49.300
Let's see, there is a vertical asymptote here, a horizontal asymptote here, and a graph approaching
00:56:49.300 --> 00:56:54.600
both the horizontal and vertical asymptotes, but never actually reaching it.
00:56:54.600 --> 00:57:03.500
And the same over here on this side; and make sure that you denote that the graph is undefined--
00:57:03.500 --> 00:57:08.400
the function is undefined--at the excluded value of x = -3.
00:57:08.400 --> 00:57:13.000
That concludes this section of Educator.com on graphing rational functions.