Give the equations of any vertical, horizontal, or oblique asympt...

Question

Give the equations of any vertical, horizontal, or oblique asymptotes for the graph of each rational function. See Example 4. ƒ(x)=(x^2-2x-3)/(2x^2-x-10)

Accepted Answer

Hey, everyone in this problem for the following function, we're asked to write the equations for the vertical, horizontal or oblique atom totes. We're given the function F of X is equal to X squared minus five, X minus 13, divided by four, X squared minus X minus 18. We have four answer choices. Option A, a vertical Asymptote at X equals negative two and a horizontal Asymptote at Y equals negative one quarter. Option B A vertical Asymptote at X equals 9/ and X equals negative two and a horizontal ascent to at Y equals one quarter. Option C A vertical Asymptote at X equals negative two and a horizontal Asymptote at Y equals one quarter. And option D A vertical Asymptote at X equals nine divided by four and X equals negative two and an oblique Asymptote at Y equals one quarter. Now we're gonna start by rewriting this function we were given. So we have F of X is equal to X squared minus five, X minus 13 divided by four, X squared minus X minus 18. And we're gonna start with horizontal asymptotes. And when we're, when we're looking to find out if we have horizontal asymptotes. And what they are, we want to look at the degree of the numerator and the degree of the denominator of our function. And so the degree of the numerator in this case is going to be equal to two. OK, remember the degree of a polynomial is the highest exponent. So we have a quadratic polynomial in the numerator. The highest exponent is two. So the degree is two. Now, in the denominator, we have the same thing, the highest exponent is two. So the degree of the denominator is also two. So we have that the degree of the numerator is equal to the degree of the denominator which tells us that we have a horizontal Asymptote and it's gonna be located where Y is the ratio of the leading coefficients. OK? So we have Y is equal to the leading coefficient of the numerator divided by the leading coefficient of the denominator. Now I've written this equation as Y equals your horizontal AY will always be an equation of Y equals. OK. If you imagine a horizontal line on a graph that has a particular Y value, the line equation for that line, a horizontal line is Y equals OK. So our horizontal AY will always be Y equals not X equals. Now, the leading coefficient of the numerator is the coefficient corresponding to that highest exponent term. The highest exponent was two. So the X squared term is our highest exponent term. And it has a coefficient of one. So we get Y is equal to one divided by the leading coefficient of the denominator. Same thing in the denominator. The highest exponent term is our X squared term. It has a coefficient of four. So we get our horizontal as tode at Y equals one divided by four. Now, because we have a horizontal Asymptote, it means that we don't have an oblique Asymptote for the horizontal aceto, we found that we had the degree of the numerator equal to the degree of the denominator which gave us a horizontal ascent tope an oblique ascent tote, we need the degree of the numerator to be one bigger than the degree of the denominator. And so any function can't satisfy both of those conditions. OK. So as soon as we have a horizontal Asymptote, it means we don't have an oblique Asymptote, which is the case with this function. So let's move to our vertical Asymptote and for our vertical as to what we wanna do is we wanna look where the denominator is equal to zero. And when I say the denominator is equal to zero, I mean, after simplify, OK. So we want to simplify as much as we can cancel any terms that we can cancel. And then look where the denominator is equal to zero. Now, our function F of X is equal to X squared minus five, X minus 13 divided by four X squared minus X minus 18. OK. How can we simplify this? Well, the numerator actually can't be simplified nicely. OK. That does not have nice integer factors. So we can't factor that or simplify any further. So we actually won't have any canceling that can happen. So let's look at the denominator now, so we have four X squared minus X minus 18 equal to zero. Now, we can break up this middle term minus X and write it as four X squared plus A X minus nine, X minus 18, equal to zero. OK? Why do we do that? It's just a way for us to factor this. And if you like to factor in a different way, then you go ahead and do that. I'm just showing you one way that you can factor this. So we've broken this middle term of minus X into eight X minus nine X. OK. Those are equivalent and now we can factor this in two parts. So out of these first two terms, four X squared plus eight X, we can factor out four X and we're gonna be left with X plus two, out of the last two terms negative nine X minus 18, we can factor negative nine. And again, what's left over is X plus two. OK. So we have four X multiplied by X plus two minus nine, multiplied by X plus two. Now we have this X plus two factor in each of these terms. We can factor that out. And what we're left with is four, X minus nine, multiplied by that factor of X plus two. And this is all equal to zero. So now we have this factor and we can look at where each of these factors is equal to zero. OK? These are being multiplied together. So if either of them is equal to zero, the entire thing will be equal to zero. So from the first factor four, X minus nine, we set that to zero and we get that X is equal to 9/4. And from the second factor we set that equal to zero, X plus two is equal to zero. And we get that X is equal to negative two. So we actually found two vertical AOM totes. We found one at X equals 9/4 and one at X equals negative two. So looking at our answer traces, we found that we had a horizontal Asymptote at Y equals 1/ and we had two vertical asymptotes, one at X equals 9/4 and one at X equals negative two. So we have answer choice. B thanks everyone for watching. I hope this video helped see you in the next one.

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Give the equations of any vertical, horizontal, or oblique asymptotes for the graph of each rational function. See Example 4. ƒ(x)=(x^2-2x-3)/(2x^2-x-10)