slant asymptote (1 Viewer)

[muttiah]

how do i do the working out for this question to find the slant asymptote

x^2- 2 / (x+5)..

how do i do the working out to find slant asymptotes.. i know its divide by highest power of x in denominator.. but do i need to put lim as x > infinity...

what i do.. thanks..
and y does the asymptote appear if i divide by hghest powere of x

 

[airie]

You could use long division or any other method you like, to get x²-2 = (x+5)(x-5) + 23, therefore (x²-2)/(x+5) = x-5 + 23/(x+5). Since the limit of this expression as x approaches +/- infinity is x-5 (as x+5 approaches +/-infinity when x approaches +/- infinity, thus 23/(x+5) approaches zero from above and below respectively), the oblique asymptote of this graph is y=x-5.

Dividing is just a method to make the oblique asymptote more apparent, it is merely rearranging the expression, nothing is changed about it. Essentially you're trying to get a constant over an expression that approaches infinity (thus the whole fraction approaches zero) as the variable approaches infinity, so the rest of the expression describes the values towards which the values of the graph approaches, as the variable approaches infinity.

 

[muttiah]

ok thanks for a good explanation.. i never knew that method.. i always divided by the highest power of x...

 

[pLuvia]

ok thanks for a good explanation.. i never knew that method.. i always divided by the highest power of x...

That's for finding horizontal asymptotes

 

[muttiah]

but in the book when i divide by thew highest power of x. i got slant or parabola asymptote and on the answers they had the same thing :S

 

[pLuvia]

but in the book when i divide by thew highest power of x. i got slant or parabola asymptote and on the answers they had the same thing :S

I thought you meant divide all terms by the highest power of x