SOLUTION: find the verical and horizontal asymptotes of the following functions: 1) f(x)=(2x^2+5x^2-3)/(x^2+2) 2) f(x)=1/(x-3)^2 thank you

Question 36677: find the verical and horizontal asymptotes of the following functions:
1) f(x)=(2x^2+5x^2-3)/(x^2+2)
2) f(x)=1/(x-3)^2
thank you
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
find the verical and horizontal asymptotes of the following functions:
1) f(x)=(2x^2+5x^2-3)/(x^2+2)
Did you mean to have two "x^2" terms?
Horizontal Asymptotes depends on your answer to that question.
If you meant 2x^2+5x-3 in the numerator, you have a horizontal
asymptote at y=2.
If you menat 2x^2+5x^2-3 = 7x^2-3 as the numerator, you have a
horizontal asymptote at y=7.
Vertical Asymptotes:
When/If the denominator is zero. But x^2+2 cannot be zero. So,
no Vertical asymptotes.
2) f(x)=1/(x-3)^2
Horzontal Asymptotes: none
Vertical Asymptotes: When (x-3)^2=0
When x=3
Therefore: Vertical asymptote at x=3
Cheers,
Stan H.

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