y = ( x^3 + 5x^2 + 3x + 10 )/( x^2 + 1 )
2007-05-22 18:09:47 · 4 answers · asked by fye 1 in Science & Mathematics ➔ Mathematics
how do you do long division? my night school class never covered it with, we used synthetic.
2007-05-22 18:13:29 · update #1
y = (x^3 + 5x^2 + 3x + 10) / (x^2 + 1)
Use synthetic long division. You're going to end up with a expression which looks something like
y = f(x)/(x^2 + 1) + g(x)
Your oblique asymptote will be y = g(x).
2007-05-22 18:13:33 · answer #1 · answered by Puggy 7 · 0⤊ 0⤋
y = ( x^3 + 5x^2 + 3x + 10 )/( x^2 + 1 )
y= x+5+(3x+5)/(x²+1)
the oblique asymptote is y=x+5
the remainder is (3x+5)/(x²+1); meaning that the difference at any point x between the curve y = ( x^3 + 5x^2 + 3x + 10 )/( x^2 + 1 ) and the line y=x+5 will be (3x+5)/(x²+1)
good luck to you
2007-05-23 01:17:22 · answer #2 · answered by Nick 2 · 0⤊ 0⤋
The most significant terms in the numerator is x^3 + 5x^2, and for the denominator it's x^2. x^3 + 5x^2 divided by x^2 gets you x+5, so that the "oblique asymptote" is y = x+5.
2007-05-23 01:20:01 · answer #3 · answered by Scythian1950 7 · 0⤊ 1⤋
do long division.. ignore any remainder because as x=>oo the fraction will get closer and closer to zero.
Division yields x+ 5 + remainder
asymptote is x+5
_____________________
remeber to put in your "place holder" so x²+1 is x² + 0x +1
in doing your long division
=]
2007-05-23 01:12:27 · answer #4 · answered by Anonymous · 0⤊ 0⤋