SOLUTION: determine all asymptotes(vertical,horizonalor oblique) f(x)= 2^2-3x+2/4x^2-1

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Question 294585: determine all asymptotes(vertical,horizonalor oblique)
f(x)= 2^2-3x+2/4x^2-1
Answer by Fombitz(32388) About Me  (Show Source):
You can put this solution on YOUR website!
Vertical asymptotes occur when the denominator goes to zero.
4x%5E2-1=0
4x%5E2=1
x%5E2=1%2F4
x=1%2F2 and x=-1%2F2
The vertical asymptotes are shown below in blue.
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Horizontal asymptotes occur as x goes to infinity.
Divide numerator and denominator by x%5E2.
f%28x%29=%282x%5E2-3x%2B2%29%2F%284x%5E2-1%29=%282-3%2Fx%2B2%2Fx%5E2%29%2F%284-1%2Fx%5E2%29
Now as x goes to infinity,
f%28x%29=%282-cross%283%2Fx%29%2Bcross%282%2Fx%5E2%29%29%2F%284-cross%281%2Fx%5E2%29%29
the crossed out terms go to zero.
lim%28+x-%3Einfinity%2C+f%28x%29+%29+%0D%0A=+2%2F4=1%2F2
The horizontal asymptote would be y=1/2 (shown below in green).
There are no oblique asymptotes.
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