SOLUTION: Find a horizontal asymptote, it it exists, of the function g(x)=(1-x)(2+3x)/2x^2+1 Help me please, I am not good with asymptotes much.

Algebra ->  Trigonometry-basics -> SOLUTION: Find a horizontal asymptote, it it exists, of the function g(x)=(1-x)(2+3x)/2x^2+1 Help me please, I am not good with asymptotes much.       Log On


   

Question 982454: Find a horizontal asymptote, it it exists, of the function
g(x)=(1-x)(2+3x)/2x^2+1
Help me please, I am not good with asymptotes much.
Found 2 solutions by Edwin McCravy, AnlytcPhil:
Answer by Edwin McCravy(20081) About Me  (Show Source):
You can put this solution on YOUR website!
I'll post it in a few minutes:

Answer by AnlytcPhil(1810) About Me  (Show Source):
You can put this solution on YOUR website!
[Edwin McCravy and AnlytcPhil are the same person!]

g%28x%29%22%22=%22%22%281-x%29%282%2B3x%29%2F%282x%5E2%2B1%29

We multiply the numerator out:

g%28x%29%22%22=%22%22%281%2Bx-3x%5E2%29%2F%282x%5E2%2B1%29

Place the numerator in descending order of powers of x:

g%28x%29%22%22=%22%22%28red%28-3%29x%5E2%2Bx%2B1%29%2F%28red%282%29x%5E2%2B1%29

Since the degree of the numerator is 2 and the degree
of the denominator is also 2, the equation of the
horizontal asymptote is 

y%22%22=%22%22red%28-3%29%2Fred%282%29

y%22%22=%22%22-3%2F2



The green line is the horizontal asymptote.  It's hard to tell but on
the left the graph actually crosses its horizontal asymptote at x= -3.5,
goes a tiny bit below it and then approaches it from below.  On the right
the graph only approaches the horizontal asymptote from above.

Edwin