In Exercises 57–64, find the vertical asymptotes, if any, the hor...

Question

In Exercises 57–64, find the vertical asymptotes, if any, the horizontal asymptote, if one exists, and the slant asymptote, if there is one, of the graph of each rational function. Then graph the rational function. g(x) = (4x^2 - 16x + 16)/(2x - 3)

Accepted Answer

Hello, everyone in this video. We're going to be looking at this practice problem where we have the rational function F of X is equal to x squared minus four X plus four divided by X plus two. And from the rational function we want to try to find the horizontal or vertical or slight ascent. Oh if there are any. So I'm going to rewrite this rational function off to the side here so I have F of X is equal to x squared minus four. X plus four divided by X plus two. So whenever we're looking at ascent, oats or rational functions in general, one of the more important features to look at is the degree of the numerator will denote D. N. And the degree of the denominator which I'll denote D. D. In this case the degree of the numerator is two and the degree of the denominator is one. So because we have the degree of the numerator greater than the degree of the denominator, I can already say that there will be no horizontal ascent and in this case the degree of the numerator is greater than the degree of the denominator by one. So that means that I have to worry about slight slant ascent. So the special condition where a slight slant a cento is present is when the degree of the numerator, it is greater than the degree of the denominator by one. And in this case that is what we have. So we're going to have to worry about a slant ascent up in order to solve for any vertical ascent. Okay we look at the denominator. So in this case we have denominator of X plus two. So if I want to solve for the vertical ascent, Oh I want to try to find the root of the denominator And to find the route. All set this expression equal to zero. And to solve for X, I'll subtract two. So that gives me a vertical ascent. OX. is equal to negative two. So we'll go ahead and write that in. So I've already determined that I have a vertical ascent. X equals negative two and no horizontal asientos. So now we can worry about the slant asientos and I can determine the slant slant a cento by dividing the numerator, X squared minus four X plus four divided by the denominator, X plus two. So I can start with this long division. So the first thing I'm going to do is look at the leading terms. I have X squared divided by X. And that leaves me with just X. So if I write X. And this first spot here and then multiply it will have minus X squared minus two X. So my new line no longer has any X squares but it now has negative six sex plus four. And once again I can look at my leading terms. So I'll do negative six sex divided by X. And that leaves me with just negative six. So that's what I will write up here and once again I'll multiply that. So now I have negative six sex And -6 times positive two is -12. If I distribute this negative, I'll have plus six six plus 12. So I'm left with zero plus 16. And because when I do 16 divided by X, I can't really simplify Much else. 16 is my remainder. So this division actually leaves me with X minus six plus 16 divided by X plus two. And for our slant ascent, Oh, we actually ignore the remainder portion and we only care about this expression here. So I can say that our slant ascent Ope is at x minus six or more specifically, Y is equal to X -6 at our slant ascent. Okay. And now we can look up the graphs and sort of start to visualize this. And just to make it easier to visualize, you can see that at the origin or 00. The ascent tote should fall below the X axis because I have zero is equal to zero minus six. Or sorry, we don't know the value of why. So when X is equal to zero will have y is equal to negative six. So, I'm looking for a graph that has the ascent below the X axis and the vertical ascent is at X to the X equals negative two, which means that I want the ascent to to be left of the Y axis. So if we look at the graphs, you can see that a has the ascent tote to the left of the y axis. So that's good. We can also see that the ascent tote line is below the X axis. So that's good. So A could be a possible answer. If we look at B, you can see that it has the ascent, the vertical ascent to the right of the X axis, which is not something we want. So B is eliminated as a potential graph. And if we look at, see once again we see the vertical ascent to the left, something that we want and the slant ascent tote below the X axis, something else that we were looking for. And once again with D. We see the same conditions where we have the vertical ascent to the left of the X axis, the y axis and the ascent below the X axis at zero X. Egal zero. So now we can look at the graphs that fulfilled our basic conditions and see which one is more specifically applicable to our rational function. And specifically this coordinate here that we solved where if We plug in zero into our rational function for X. So if I plug in zero, all we left with why is equal to four divided by two or just two. So if I have X0, Y should be two. So we're gonna use this point to figure out which of these graphs specifically apply to our rational function. And here you can see that we have 04, which is not what we want, so A. Is eliminated. We already eliminated B. And you can see this point is at 0 -2, which is not what we want. So see is eliminated. And finally we have zero and 2, which is what we're looking for. So you can see that D. Best follows our rational function, so D. Is our final answer. So hopefully this video helped and see you next time.

In Exercises 57–64, find the vertical asymptotes, if any, the horizontal asymptote, if one exists, and the slant asymptote, if there is one, of the graph of each rational function. Then graph the rational function. g(x) = (4x^2 - 16x + 16)/(2x - 3)

Key Concepts

Vertical Asymptotes

Horizontal and Slant Asymptotes

Graphing Rational Functions

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