# What is the slant asymptote of the function y=x^2/(x+1)?

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### 2 Answers

y=x^2/(x+1) .

We know that x^2 divided by (x+1) gives a quotient (x-1) and a remainder 1.

=> x^2/(x+1) = (x-1)+1/(x+1)

Therefore y = (x-1) +1/(x+1)

Therefore Lim x-> inf y = Limx-> inf {(x-1) + 1/(x+1)} but lim x-> inf 1/(x+1) = 0

Therefore y = x^2/(x+1) has the asymptote** y = x-1**.

We'll recall the equation of the slant asymptote:

y = mx + n

We must identify the coefficients m and n to determine the equation of the slant asymptote.

m = lim f(x)/x, if x approaches to +infinite

Let f(x)=y

m = lim x^2/x*(x+1)

We'll remove the brackets from denominator:

lim x^2/x*(x+1) = lim x^2/(x^2 + x)

We'll force the factor x^2 at denominator:

lim x^2/x^2*(1 + 1/x) = lim 1/(1 + 1/x)

lim 1/(1 + 1/x) = lim 1/(1 + lim 1/x)

lim 1/(1 + lim 1/x) = 1/(1+0) = 1

Since m = 1, we may calculate n:

n = lim [f(x) - mx] = lim [x^2/(x+1) - x]

lim [x^2/(x+1) - x] = lim (x^2 - x^2 - x)/(x+1)

lim (- x)/(x+1) = -1/1 = -1

**The equation of the slant asymptote, if x approaches to + infinite and - infinite, is y = x - 1.**