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^2-1)/(x+3)

Accepted Answer

Hey, everyone in this problem for the following function, we're asked to write the equations for the vertical, horizontal or oblique asom totes. The function for given is F of X is equal to X squared minus nine divided by X plus 17. We're given four answer choices. Option A through D option A B and C show different equations for the vertical Asymptote. And for an oblique Asymptote option D has an equation for a vertical Asymptote and a horizontal Asymptote. We're gonna start by rewriting the function we were given. So we have F of X is equal to X squared minus nine divided by X plus 17. Let's start with our horizontal asymptotes. OK. When we're looking at horizontal asymptotes, we wanna compare the degree of the numerator and the degree of the denominator of our function. So the degree of the numerator in this case is equal to two. OK. The degree, remember of a polynomial is the highest exponent that you have. The highest exponent is two. So the degree is two, we also have the degree of the denominator and again, the highest exponent. And in this case, we have X plus 17, the highest exponent is one. And so we have the degree of the denominator is just one. And what you'll notice is that the degree of the numerator is greater than the degree of the denominator. OK. This tells us that we have no horizontal Asymptote. OK? For a horizontal Asymptote, we need the degree of the numerator and denominator to either be equal or for the degree of the denominator to be greater than that of the numerator. OK. So we have no horizontal Asymptote. That means we can already eliminate answer choice. D OK. It has a horizontal Asymptote in the second part of that answer choice. Um And we know we don't have one now, we can move to our vertical ascent to and the vertical as is going to occur where the denominator is equal to zero. OK? And I say that meaning after simplifying, so we want to simplify the function as much as we can. We want to factor, we wanna cancel any terms and then we want to set the denominator equal to zero. Now, in our function, we can factor that numerator. OK. This is a difference of squares. So we'll get X minus three multiplied by X plus three, but those factors aren't going to cancel with anything. There's gonna be no more simplifying. And so we're gonna just take that denominator and set it equal to zero. So we have X plus 17 is equal to zero which tells us that X is equal to negative 17. OK. So we have a vertical ascent to X is equal to negative 17. And again, looking at our answer traces, we can eliminate another one. We can eliminate C in this case because it has a vertical ascent to at X equals positive 17. Or we found it at X equals negative 17. All right. Horizontal Aceto, its done vertical ascent tode is done moving on to our oblique asymptotes. Now, for our oblique as totes and what we want to notice is that the degree of the numerator that we found, which was two is equal to the degree of the denominator plus one. OK. So the degree of the numerator is one more than the degree of the denominator. Remember we had the degree of the numerator is two. The degree of the denominator is one. So it satisfies this condition which tells us that we have and gas them to him. Now, how do we find it? Well, what we're gonna do is we're gonna take the long division, we're gonna divide this function, the numerator by the denominator. You gonna draw our long division line here underneath it. We're gonna write the numerator. In this case, it's X squared plus zero X minus nine. OK? I included that plus zero X. It's very, very important you need to include every single term, even the ones that have coefficients of zero. OK. So you have to work down from your exponents, we had an exponent of two. Then we need to do our exponent one term and then our constant term. All right. And then on the left, we're gonna write out the denominator, the thing that we're dividing by which is X plus 17. And now we can work through our division. So how we're gonna do this is we're gonna take this first value under our line X squared. We're gonna divide it by the first term in our denominator, which is X X squared divided by X gives us X. And we write that up top. Now we're gonna take that value. We found up top and multiply it by the entire denominate, multiplied by X plus 17 gives us X squared plus 17 X. And we write it underneath that numerator. OK? So we have X squared plus 17 X and then we're gonna subtract within these columns. OK? So we have X squared minus X squared gives us zero. We have zero minus 17 X gives us negative X. OK? So you can see we've now eliminated the X word terms and this is kind of how long division works. We work through these terms one at a time until we get down to just a constant remainder left. All right. So now we have negative 17 X and we're gonna do the same thing. We're gonna divide it by the first term of the denominate negative 17 X divided by X gives us negative 17. We write it up above. We take that value of negative 17. We multiply it to the entire denominator of X plus 17. This gives us negative 17 acts minus 289. Now we need to bring this negative nine all the way down into this section. And we're gonna subtract these columns again. So we have negative 17 X minus negative 17 X. This gives us zero and then we have negative nine minus negative 289 which gives us 280. So we're done our long division. We need to interpret the results. Now, the oblique Asymptote remember is going to have the equation of a line. And so our oblique Asymptote is going to be at Y equals and it's gonna be at Y equals the coffin of the division. And the quo is the value the term that we get above our division symbol. So in this case, that's X minus 17. And so we have our oblique Asymptote, Y equals X minus 70. And now we've done all three pieces. We found that there was no horizontal Asymptote. We found that there was a vertical Asymptote at X equals negative 17 and an oblique Asymptote at Y equals X minus 17. If we compare this with our answer choices, we see that we have answer choice. B thanks everyone for watching. I hope this video see you in the next one.

Struggling with College Algebra?

Give the equations of any vertical, horizontal, or oblique asymptotes for the graph of each rational function. See Example 4. ƒ(x)=(x^2-1)/(x+3)

Watch next

Asymptotes practice set