1. ## Oblique Asymptotes

hi , I learned in class that if an Oblique asymptote is present, then the degree of the numerator should be exactly 1 more than the denominator's. However this does not apply to $f(x) = (x^3-16x)/(-4x^2+4x+24)$ how come?

I mean I looked at its appropriate graph, and I could not see an OA, According to my calculations it is: $y= 1/4 x + 1/4$.

2. ## Re: Oblique Asymptotes

Originally Posted by sakonpure6
hi , I learned in class that if an Oblique asymptote is present, then the degree of the numerator should be exactly 1 more than the denominator's. However this does not apply to $f(x) = (x^3-16x)/(-4x^2+4x+24)$ how come?

I mean I looked at its appropriate graph, and I could not see an OA, According to my calculations it is: $y= 1/4 x + 1/4$.
You are off by a - sign. See below.

-Dan

3. ## Re: Oblique Asymptotes

Oh okay! By the way on the right hand side of the graph, does the function approach the OA or get farther from it? Because to me it looks like it is getting farther.

4. ## Re: Oblique Asymptotes

Hello, sakonpure6!

$\text{Find the oblique asymptote: }\:f(x) \:=\: \frac{x^3-16x}{-4x^2+4x+24}$

$\text{We have: }\:f(x) \;=\;\frac{x^3-6x}{\text{-}4x^2 + 4x + 24}$

$\text{Divide numerator and denominator by }x^2\!:$

. . $f(x) \;=\;\dfrac{\frac{x^3}{x^2} - \frac{6x}{x^2}}{\frac{\text{-}4x^2}{x^2} + \frac{4x}{x^2} + \frac{24}{x^2}} \;=\;\frac{x - \frac{16}{x}}{\text{-}4 + \frac{4}{x} + \frac{24}{x^2}}$

$\text{Hence: }\:\lim_{x\to\inty}f(x) \;=\;\lim_{x\to\infty} \frac{x - \frac{16}{x}}{\text{-}4 + \frac{4}{x} + \frac{24}{x^2}} \;=\;\frac{x-0}{\text{-}4+0+0} \;=\;\frac{x}{\text{-}4}$

$\text{The oblique asymptote is: }\:y \:=\:\text{-}\tfrac{1}{4}x$

5. ## Re: Oblique Asymptotes

Originally Posted by sakonpure6
Oh okay! By the way on the right hand side of the graph, does the function approach the OA or get farther from it? Because to me it looks like it is getting farther.
It lines up. Graph it on a calculator and zoom out. (Or you could do it for real and find the limit. )

-Dan

6. ## Re: Oblique Asymptotes

The oblique asymptote is y=-x/4 -1/4

7. ## Re: Oblique Asymptotes

Originally Posted by Soroban
Hello, sakonpure6!

$\text{We have: }\:f(x) \;=\;\frac{x^3-6x}{\text{-}4x^2 + 4x + 24}$

$\text{Divide numerator and denominator by }x^2\!:$

. . $f(x) \;=\;\dfrac{\frac{x^3}{x^2} - \frac{6x}{x^2}}{\frac{\text{-}4x^2}{x^2} + \frac{4x}{x^2} + \frac{24}{x^2}} \;=\;\frac{x - \frac{16}{x}}{\text{-}4 + \frac{4}{x} + \frac{24}{x^2}}$

$\text{Hence: }\:\lim_{x\to\inty}f(x) \;=\;\lim_{x\to\infty} \frac{x - \frac{16}{x}}{\text{-}4 + \frac{4}{x} + \frac{24}{x^2}} \;=\;\frac{x-0}{\text{-}4+0+0} \;=\;\frac{x}{\text{-}4}$

$\text{The oblique asymptote is: }\:y \:=\:\text{-}\tfrac{1}{4}x$
You are looking at the behavior of the function in the asymtotic region. You suppressed what is happening near the origin. Just divide the polynomials and you will get -x/4 - 1/4. The complete graph is attached.

8. ## Re: Oblique Asymptotes

Originally Posted by topsquark
You are off by a - sign. See below.

-Dan
What graphing utility you are using

9. ## Re: Oblique Asymptotes

Originally Posted by votan
What graphing utility you are using
Graph. It's freeware and it's pretty awesome, really. No pesky ads or anything.

-Dan

10. ## Re: Oblique Asymptotes

Originally Posted by topsquark
Graph. It's freeware and it's pretty awesome, really. No pesky ads or anything.

-Dan
I am using dplot95, freeware from uncle sam. It is superior. The reason I asked is because your graph seems to cross the asymptote near about x=6, suggesting there is a point of inflection somewhere. Is there one?

11. ## Re: Oblique Asymptotes

Originally Posted by votan
I am using dplot95, freeware from uncle sam. It is superior. The reason I asked is because your graph seems to cross the asymptote near about x=6, suggesting there is a point of inflection somewhere. Is there one?
I don't think there is anything to be read into the asymptote cutting across the function. It's about the limiting properties of the function at large |x|, not in the region "near" the origin.

I'll look into dplot, thanks.

-Dan

12. ## Re: Oblique Asymptotes

Originally Posted by Soroban
Hello, sakonpure6!

$\text{We have: }\:f(x) \;=\;\frac{x^3-6x}{\text{-}4x^2 + 4x + 24}$

$\text{Divide numerator and denominator by }x^2\!:$

. . $f(x) \;=\;\dfrac{\frac{x^3}{x^2} - \frac{6x}{x^2}}{\frac{\text{-}4x^2}{x^2} + \frac{4x}{x^2} + \frac{24}{x^2}} \;=\;\frac{x - \frac{16}{x}}{\text{-}4 + \frac{4}{x} + \frac{24}{x^2}}$

$\text{Hence: }\:\lim_{x\to\inty}f(x) \;=\;\lim_{x\to\infty} \frac{x - \frac{16}{x}}{\text{-}4 + \frac{4}{x} + \frac{24}{x^2}} \;=\;\frac{x-0}{\text{-}4+0+0} \;=\;\frac{x}{\text{-}4}$

$\text{The oblique asymptote is: }\:y \:=\:\text{-}\tfrac{1}{4}x$
Another way to see this is to use "long division": $\frac{x^3- 16x}{-4x^2+ 4x+ 24}= -\frac{1}{4}x- \frac{1}{4} - \frac{9x- 6}{-4x^2+ 4x+ 24}$.
Obviously as x goes to infinity (or negative infinity) the remaining fraction goes to 0 so the graph approaches $-\frac{1}{4}x- \frac{1}{4}$.

13. ## Re: Oblique Asymptotes

Originally Posted by topsquark
I don't think there is anything to be read into the asymptote cutting across the function. It's about the limiting properties of the function at large |x|, not in the region "near" the origin.

I'll look into dplot, thanks.

-Dan
Can we insert a power point presentation here?