Now divide by .
Yes, since the remaining fraction has lower degree in the numerator than in the denominator, it will go to 0 and so f(x) will go to -15.2. Find the end-behavior asymptote(s) in the graph of f(x)=(4x-15x^2)/(x^2-121).
This is how I solved for it:
I'm not sure if I did it right. I just followed example problems.
No, you did not divide correctly. Since the numerator is of degree 2 and the denominator of degree 1, you would expect a quotient of degree 2-1= 1. x-3 divides into with quotient x+ 3 and remainder 4: y= x+ 3 + 4/(x-3). The (slant) asymptote is y= x+ 3.3. Same for f(x)=(x^2-5)/(x-3).
My answer this time was y=4+((x+4)/(x-3)), so if I did the last problem correctly, the asymptote should be y= 4.
You appear to have divided incorrectly again. Could you show exactly how you got that?4. Find the polynomial end-behavior asymptote in the graph of f(x)=(2x^3+6x-1)/(x+2).
My answer was y=2x^2-4x+14
If you could explain the process, that would help a lot. Thanks!