Question

What are the asymptote(s) and hole(s), if any, of `f(x)=(x^2-1 )/(x^2+3x-4)`?

Answer 1

V.A at `x=-4` ; H.A at `y=1` ; Hole is at `(1,2/5)`

Explanation:

`f(x)= (x^2-1)/(x^2+3x-4)= ((x+1)(x-1))/((x+4)(x-1))=(x+1)/(x+4) :.`Vertical asymptote is at `x+4=0 or x=-4`; Since degrees of numerator and denominator are same, horizontal asymptote is at (numerator's leading coefficient/denominator's leading coefficient)`:.y=1/1=1`.There is a cancellation of `(x-1)` in the equation. so hole is at `x-1=0 or x=1` When `x=1; f(x)=(1+1)/(1+4)=2/5 :.` The hole is at `(1,2/5)` graph{(x^2-1)/(x^2+3x-4) [-40, 40, -20, 20]}[Ans]