f(x)=X^2+1/X^2 -1findx,y int,where are vertical& horizontal asymptotes?

Question

f(x)= x^2+1/x^2-1 find x,y intercepts if any,where are the vertical asymptotes?where is the horizontal asymptote? Solve rational equations:

a) 1/x-2+1/x+2=4/x^2-4

b) -36/x^2+x+-20 - 2/x-4 = 4/x+5

c) 7/x+5= -6/x^2

d) 3/x+4=4/x-1

Solve:

find x when f(x) = 7/(x+4)=3.

Answer 1

f(x)=(x²+1)/(x²-1); x²-1=0 when (x-1)(x+1)=0, that is, when x=1 or -1 (these are the vertical asymptotes).

f(0) is the y-intercept: 1/-1=-1; f(x)=0 would be the x-intercept, but x²+1 cannot be zero so there is no x-intercept.

The horizontal asymptote is found by replacing x with a very large number, X, so f(X)≈X²/X²=1, making the horizontal asymptote y=1.