SOLUTION: I'm not able to figure these out, can anyone help me out? Find the equations of the horizontal and vertical asymptotes for the following. Type none if the function does not hav

Algebra ->  Human-and-algebraic-language -> SOLUTION: I'm not able to figure these out, can anyone help me out? Find the equations of the horizontal and vertical asymptotes for the following. Type none if the function does not hav      Log On


   

Question 125464: I'm not able to figure these out, can anyone help me out?
Find the equations of the horizontal and vertical asymptotes for the following. Type none if the function does not have an asymptote.
a) f%28x%29=2x%2B3%2Fx%2B2 this is not showing correctly on the preview, it's supossed to be: "f(x)= 2x+3 OVER x+2".
Answer:
Horizontal:
Vertical:
b) g%28x%29=5x%2Fx%5E2%2B1 this is not showing correctly on the preview, its supossed to be: "g(x)= 5x OVER x^2+1".
Answer:
Horizontal:
Vertical:
Answer by uma(370) About Me  (Show Source):
You can put this solution on YOUR website!
(a) given f(x) = (2x+3)/(x+2)
To get the vertical asymptote, set the denominator to 0
==> x+2 = 0
==> x = -2 is the equation of the vertical asymptote.
Now divide all the terms of the function by x and set lt x-> infinity.
==> f(x) = (2 + 3/x)/ (1 + 2/x)
AS x -> infinity, we have f(x) = 2/1 = 2
Thus y = 1 is the equation of the horizontal asymptote.
(b)As x^2 + 1 = 0 does not have a real solution, there is no vertical asymptote.
Divide each term by x^2 and setting lt x-> 0, we get y = 0
So y = 0 is the equation of the horizontal asymptote.
good luck!!!