[SPLIT, MOVED] R(x)=6x^2 + x +12 divided by 3x^2 - 5x -2

[VHuhges]

I am having problems with asymptotes. I have to determine the vertical, horizontal, and oblique asymptotes for the following problem

R(x)=6x^2 + x +12 divided by 3x^2 - 5x -2


Anybody got any suggestions. I know from my text that since the degree of the numerator is equal to the degree of the numerator, you have to divide the numerator by the denominator. I just am not sure of the answer. When I do the long division I get as the answer 2 + 11x + 16 over 3x^2 -5x-2. So what is the horizontal asymptote. y=2 or y= 2 + 11x + 16 over 3x^2 -5x-2 ? This mess is confusing.

 

[mmm4444bot]

... I have to determine the vertical, horizontal, and oblique asymptotes ...

In general, rational functions have a horizontal asymptote OR an oblique asymptote; not both.

(It's possible to create an exception to this, however, by defining a piecewise function.)

R(x) = 6x^2 + x + 12 divided by 3x^2 - 5x - 2

... I know from my text that since the degree of the numerator is equal to the degree of the numerator, you have to divide the numerator by the denominator ...

Hello V:

In your future requests for help, please click the NEWTOPIC button to start your own thread; you've posted your question within somebody else's conversation, here.

I think that maybe you should look at your book, again. When the degrees are equal, we do not need to carry out the division.

We know there is no slant asymptote here. A slant asymptote can only occur when the degree of the polynomial in the numerator is one more than the degree of the polynomial in the denominator.

There are two ways to find the equation of the horizontal asymptote; one is a shortcut, and the other requires some extra work.

The shortcut is to remember that when the degrees are equal, the equation of the horizontal asymptote is y = A/B, where A is the leading coefficient in the numerator and B is the leading coefficient in the denominator.

The other method involves dividing both the numerator and the denominator by the largest power of x, which in this function is x^2.

\(\displaystyle \frac{\frac{6x^2 + x + 12}{x^2}}{\frac{3x^2 - 5x - 2}{x^2}}\)

Simplify this expression, and then let x go to infinity to see which terms go to zero; you find the expression for the horizontal asymptote with what's left over.

Either method gives you with the same equation for the horizontal asymptote. Personally, I like to use the shortcut rules for horizontal/slant asymptotes, but the division above -- followed by some analysis of what happens when x gets very big positively or negatively -- shows us why the shortcut works.

Finally, to find equations for any vertical asymptotes, you need to factor the denominator.

Vertical asymptotes occur at values of x that cause the denominator to become zero.

If you need more help with this exercise, please show us what you've been able to do, so far.

Cheers,

~ Mark

 

[VHuhges]

Thanks for your help on this. I am new to the site and I couldn't figure out how to post my question. Secondly this is a College algebra class that I am taking on-line. Thirdly I did check the book over and over again. In one of the examples where the degrees are equal in the numerator and the denominator the book says you will have to do long division to find the horizontal or the oblique asymptotes. All of the examples they are giving me in my book show using long division. Even the solutions manual shows that you use long division. Thanks for your assistance. What you have said has helped.

Thanks
Vicky

 

[mmm4444bot]

... where the degrees are equal in the numerator and the denominator the book says you will have to do long division to find the horizontal or the oblique asymptotes ...

Hello Vicky:

In the case where the degrees are equal, there is never a slant asymptote, and, regarding the horizontal asymptote's equation, that instruction seems like overkill, to me.

Depending upon the coefficients, longhand polynomial division can become labor intensive -- sometimes a little, sometimes a lot. Here's an example where the labor is "a little" intensive.

\(\displaystyle \frac{28x^2 - 13x - 6}{12x^2 - 7x + 1}\)

If we divide this algebraic fraction using longhand, we see that 12x^2 goes into 28x^2 seven-thirds times.

Multiplying the divisor by 7/3 gives 28x^2 - (49/3)x + (7/3).

Now we need to subtract this result from the numerator, which yields the remainder (10/3)x - (25/3).

\(\displaystyle \frac{28x^2 - 13x - 6}{12x^2 - 7x + 1} \; = \; \frac{7}{3} + \frac{\frac{10}{3}x - \frac{25}{3}}{12x^2 - 7x + 1}\)

I suppose that work wasn't so bad, but is it really necessary? Our goal is to find the equation of the horizontal asymptote; the work that we did to find the coefficents of 10/3 and -25/3 is certainly not necessary because knowing those values adds nothing to our knowledge of the horizontal asymptote. They are not used at all.

Continuing after this longhand division, we look at the algebraic fraction in the result and see that the numerator has degree 1 and the denominator has degree 2. So, as x goes to positive or negative infinity, this algebraic fraction goes to zero, and we're left with 7/3.

The equation of the horizontal asymptote is y = 7/3.

I think that it is easier to look at the ratio of leading coefficients in the original algebraic fraction and see 28/12, which simplifies to 7/3, and then write y = 7/3.

Even if I were not to use this "shortcut" rule, I still would not do polynomial division. I would divide top and bottom by x^2 (as I showed you with a different example in my first response), and quickly see that, as x goes to positive or negative infinity, everything goes to zero except the 28 and the 12. Therefore, again, I would get the ratio 28/12 and simplify to 7/3.

Perhaps, your book is filled with "easy" examples, so all of those longhand divisions are fairly trivial.

In cases where the degree of the numerator is one more than the degree of the denominator, then there is a slant asymptote, and longhand division is used to find its equation.

When the degree of the numerator is more than one greater than the degree of the denominator, then there is no horizontal or slant asymptote at all.

When the degree of the numerator is less than the degree of the denominator, there is a horizontal asymptote, and it's equation is always y = 0.

If you search the Internet, then I'm sure that you will find these four rules and many examples of their use. Of course, if you enjoy longhand polynomial division, then you can skip the rules, but -- as I said before -- I think it is overkill to do so in the case where the degrees are equal.

Cheers,

~ Mark

 

[VHuhges]

Thanks Mark. And no I would rather not do long hand either. It is just hard when you are taking a course like this on-line and the instructor is not available in the middle of the night to ask questions and you can only go by what the book tells you. She is a great instructor. I have called on her during the day and have stopped by campus to see her.The book came with a CD and it has a man doing odd problems. Some times he doesn't explain the steps enough. You have been very helpful. Thanks a lot. I went back over my work using the information you gave me and it helped me see where I had gotten off track, plus it helped me with some problems I hadn't tackled yet.


Thanks
Vicky