asymptotes for (2x^5 - 3x^2 + 5)/(3x^4 + 5x - 4)

[Guest]

Hi I need help with asymptotes:

1. I have to find the horizontal asymps. for f(x)= [2x^5-3x^2+5]/ [3x^4+5x-4]
when x-> oo I got 66 and for x-> -oo I got -66 so the answer is suppose to be oo, does that mean when theres just a high number it approaches oo?

Also, y=(x^2-x-6)/(x-3) [you have to state the vertical asym, and where it discontinues]

It discontinues at 3, so x can not equal 3, I thought that means the asymp. would be 3 too but the answer is that it doesn't have an asym. Why is this?

and how do you find the vertical asym for y=xlnx? and where it discontinues?

thanks for the help

 

[galactus]

Re: ASYMPTOTES

Hi I need help with asymptotes:

1. I have to find the horizontal asymps. for f(x)= [2x^5-3x^2+5]/ [3x^4+5x-4]
when x-> oo I got 66 and for x-> -oo I got -66 so the answer is suppose to be oo, does that mean when theres just a high number it approaches oo?

If the degree of the numerator is greater than the degree of the denominator(5>4), then there is no asymptote because f(x) approaches infinity as x approaches infinity

Also, y=(x^2-x-6)/(x-3) [you have to state the vertical asym, and where it discontinues]

It discontinues at 3, so x can not equal 3, I thought that means the asymp. would be 3 too but the answer is that it doesn't have an asym. Why is this?

Factor: \(\displaystyle \L\\x^{2}-x-6=(x-3)(x+2)\)

So, you have: \(\displaystyle \L\\\frac{(x-3)(x+2)}{x-3}=x+2\)

Now, you see why?.

and how do you find the vertical asym for y=xlnx? and where it discontinues?

ln(0) is undefined, a no-no.

thanks for the help

 

[soroban]

Hello, bittersweet!

1. I have to find the horizontal asymptotes for \(\displaystyle \L\,f(x)\:=\:\frac{2x^5\,-\.3x^2\.+\,5}{3x^4\,+\,5x\,-\,4}\)

When \(\displaystyle x\,\to\,\infty\), I got 66 . . . How?

You can "eyeball" the function as Galactus suggested,
\(\displaystyle \;\;\) or you can take the limit like this . . .

Divide top and bottom by \(\displaystyle x^4:\L\;\;f(x)\:=\:\frac{2x\,-\,\frac{3}{x^2}\,+\,\frac{5}{x^4}}{3\,+\,\frac{5}{x^3}\,-\,\frac{4}{x^4}}\)

Then: \(\displaystyle \L\,\lim_{x\to\infty}f(x)\;=\;\lim_{x\to\infty}\left(\frac{2x\,-\,\frac{3}{x^2}\,+\,\frac{5}{x^4}}{3\,+\,\frac{5}{x^3}\,-\,\frac{4}{x^4}}\right) \;= \;\frac{\infty\,-\,0\,+\,0}{3\,+\,0\,-\,0} \;=\;\infty\)

See? . . . There is no horizontal asymptote.

 

[Guest]

When \(\displaystyle x\,\to\,\infty\), I got 66

How?

I plugged in a big number like 1000 or 100, I forget which. Oh okay i'll just follow the power rule, if the power is greater than the bottom than there is no asymp.

 

[skeeter]

... okay i'll just follow the power rule, if the power is greater than the bottom than there is no horizontal asymp.

there may be oblique aymptotes, quadratic asymptotes, etc.