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Maths, asymptotes

Could somebody tell me how to work out the asymptotes for :

y = (2x² + x - 15) / (x - 3)

thanks.
Reply 1
Avatar for ShOcKzZ
ShOcKzZ
OP
7
ShOcKzZ
Could somebody tell me how to work out the asymptotes for :

y = (2x² + x - 15) / (x - 3)

thanks.

ahem
ShOcKzZ
ahem

firstly from the expression if x was 3 then the denominator would be zero,which cannot be so x=3 is a asymptote.
to work out the other asymptote use long division,and let x approach infinity.
Reply 3
Avatar for ShOcKzZ
ShOcKzZ
OP
7
IntegralAnomaly
firstly from the expression if x was 3 then the denominator would be zero,which cannot be so x=3 is a asymptote.
to work out the other asymptote use long division,and let x approach infinity.

yeh i knew about the vertical asymptote.
can u show me how to do the 2nd bit?
Reply 4
Avatar for Bezza
Bezza
2
ShOcKzZ
Could somebody tell me how to work out the asymptotes for :

y = (2x² + x - 15) / (x - 3)

thanks.

To work out the other asymptote you can divide all terms in the fraction by x to get y = (2x + 1 - 15/x)/(1-3/x) As x tends to positive or negative infinity, 15/x and 3/x tend to 0 so the curve tends to y= 2x + 1
use long division to obtain 2x+7+6/(x-3).
Now is x becomes sufficiently large 6/(x-3) approaches zero and the curve approaches 2x+7,so 2x+7 is the other asymptote.
Reply 6
Avatar for Bezza
Bezza
2
IntegralAnomaly
use long division to obtain 2x+7+6/(x-3).
Now is x becomes sufficiently large 6/(x-3) approaches zero and the curve approaches 2x+7,so 2x+7 is the other asymptote.

Do you know what's wrong with the way I tried to do it? (apart from it giving the wrong answer!)
Reply 7
Avatar for ShOcKzZ
ShOcKzZ
OP
7
hmm i dont understand this:

i once asked for the asymptotes for:

y = (x+3)/(2x-3)

ok so i was told that u divide the numerator by x to get 1+3/x
then u divide the denominator by x to get 2-3/x

so u have (1 + 3/x) / ( 2 - 3/x) x's cancel out so u have 1/2 - which is correct.

so y dnt u use the same method here? :confused:
Bezza
Do you know what's wrong with the way I tried to do it? (apart from it giving the wrong answer!)

i Can't see the wrong doing in ur method!
IntegralAnomaly
i Can't see the wrong doing in ur method!

instead of long division why dont u just put x=500, and x = 10000
IntegralAnomaly
i Can't see the wrong doing in ur method!

instead of long division why dont u just put x=500, and x = 10000.
hey ne one got p4 on the 29th?
mathematician
instead of long division why dont u just put x=500, and x = 10000.
hey ne one got p4 on the 29th?

why would u want to put x=500 and x=10000?
i've got p4 on the 29th!

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