How to find oblique asymptote? (1 Viewer)

[Sien]

P(x)=D(x)Q(x)=R(x)
therefore: y= P(x)/D(x)=Q(x)+R(x)/D(x)
Can someone please elaborate on this formula? I was away for a lesson so i just basically copied down my friend's notes which aren't even complete.
eg. (x^2 + 4x - 5)/(x+3)
How would i use that formula to work out the oblique asymptote?

 

[VBN2470]

The first formula you stated is just the transformation of a polynomial into a product of two lower degree polynomials (the divisor D and quotient Q polynomials) plus a remainder polynomial. Note that deg(R) < deg(D). To answer your question, if you want to sketch your function you can decompose the numerator into the form D(x)Q(x) + R(x) and divide by D(x) (which is essentially polynomial long division you learn in 3U). So you would have:

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Check to see if it's equal to the original function f. Your oblique asymptote will then be y = x + 1 since for very large values of x (both positive and 'largely' negative), f will tend to x + 1 (can check this graphically too).