# Complex Oblique Asymptote problem

• Jan 17th 2009, 02:36 PM
Kipper
Complex Oblique Asymptote problem
Hello,

I'm currently taking Grade 12 advance functions math and our teacher assigned an interesting question for us to work through over the weekend.

I have to graph the following equation which includes an Oblique Asymptote:

X^3 + 4x^2 + x - 6 / x^2 + 3x + 2

The first thing I did was factor the denominator and received (X+1) (X+2).

I believe I now have to do long division to determine the roots for the equation in the numerator. I was thinking about using:

X^3 + 4x^2 + x - 6 / X+1 but when I do I come up with the answer x^2 + 3x - 2.

Can someone offer some assistance on how to tackle this problem? I've used my graphing calculator to see how the graph looks and there is also a 'hole' in it as well.

Any assistance would be appreciate because I really have no idea where to go from here.

Thanks
• Jan 17th 2009, 02:53 PM
vincisonfire
First you can see that the numerator has a solution at x =1.
$x^3 + 4x^2 + x - 6 = (x-1)(x^2+5x+6) = (x-1)(x+2)(x+3)$
Now you can simplify a bit your function
$f(x) = \frac{(x-1)(x+2)(x+3)}{(x+1)(x+2)} = \frac{(x-1)(x+3)}{(x+1)}$
You must mention that the function is not defined at x=-2 though and keep that in mind.
You should see that around -1 there will be a vertical asymptote since you divide by $(x+1)$.
Then you must check what is the long term behavior. If you know the rudiment of calculus this is easy, take de derivative and check for its value as $x \rightarrow \infty$
Else, you could argue that the higher order term at the top and the bottom have the same coefficient (1) giving you a slope of 1 for the oblique asymptote. The lower order term become less and less significant as x goes up (of dowm to negative infinity).
You can also take two very high value of x (e.g. $10^7 \text{ and } 2 \cdot 10^7$ altough in your case 200 would do the job) and take the slope between them. You can take a third point and do the same thing to be sure that the function is really constantly going up at these values.
• Jan 17th 2009, 03:23 PM
mr fantastic
Quote:

Originally Posted by Kipper
Hello,

I'm currently taking Grade 12 advance functions math and our teacher assigned an interesting question for us to work through over the weekend.

I have to graph the following equation which includes an Oblique Asymptote:

X^3 + 4x^2 + x - 6 / x^2 + 3x + 2

The first thing I did was factor the denominator and received (X+1) (X+2).

I believe I now have to do long division to determine the roots for the equation in the numerator. I was thinking about using:

X^3 + 4x^2 + x - 6 / X+1 but when I do I come up with the answer x^2 + 3x - 2.

Can someone offer some assistance on how to tackle this problem? I've used my graphing calculator to see how the graph looks and there is also a 'hole' in it as well.

Any assistance would be appreciate because I really have no idea where to go from here.

Thanks

Since $y = \frac{x^3 + 4x^2 + x - 6}{x^2 + 3x + 2} = \frac{(x-1)(x+2)(x+3)}{(x+1)(x+2)}$ it should be clear that there's a hole at $x = -2$.

For $x \neq -2$ you have:

$\frac{(x-1)(x+3)}{x+1} = \frac{x^2 + 2x + 3}{x+1} = x + 1 + \frac{2}{x+1}$

and it should be clear that the equation of the oblique asymptote is $y = x + 1$ and the equation of the vertical asymptote is $x = -1$.

You should be able to find the axes intercepts. Finding exact x-coordinates of stationary points are found using calculus (and you may not have studied this to a suffiicent level).
• Jan 17th 2009, 04:34 PM
Kipper
Mr. Fantastic...

How were you able to determine that the hole is at X= -2?

We really haven't ventured into calculus at all and have really only started looking at Oblique asymptotes so I'm really not well versed on all the ways that this can be handled.

Thanks again for your help...it's been really great!!
• Jan 17th 2009, 05:27 PM
mr fantastic
Quote:

Originally Posted by Kipper
Mr. Fantastic...

How were you able to determine that the hole is at X= -2?

We really haven't ventured into calculus at all and have really only started looking at Oblique asymptotes so I'm really not well versed on all the ways that this can be handled.

Thanks again for your help...it's been really great!!

What happens to the function at x = 2 ....?
• Jan 18th 2009, 04:52 AM
stapel
Quote:

Originally Posted by Kipper
How were you able to determine that the hole is at X= -2?

Any time a variable factor, such as $x + 2$, can be "cancelled", you will get a "hole" like this. In effect, that factor served only to give you a division-by-zero problem. Otherwise, it had no effect. So the only (visible) effect of that factor in the graph will be the hole where it would otherwise have caused a problem.

Have fun! :D