# Find the slant asymptote of f(x)=x^3/(x+2)^2

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### 2 Answers

To find the slant asymptote of x^3 / (x + 2)^2 we have to divide x^3 by (x + 2)^2

(x^2 + 4x + 4) | x^3...........................................| x - 4

...........................x^3 + 4x^2 + 4x

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.....................................- 4x^2 - 4x

..................................... - 4x^2 - 4x - 16

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...........................................................16

The required asymptote is obtained from the the quotient that is obtained from the division Here, using the same we get y = x - 4.

**The required asymptote if y = x - 4**

The equation of the oblique (slant) asymptote is:

y = mx + n

We need to determine m and n to find the equation of the slant asymptote.

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

m = lim x^3/x*(x+2)^2

We'll expand the square from denominator:

lim x^3/x*(x+2)^2 = lim x^3/x(x^2 + 4x + 4)

lim x^3/x(x^2 + 4x + 4) = lim x^3/(x^3 + 4x^2 + 4x)

We'll force the factor x^3 at denominator:

lim x^3/x^3(1 + 4/x + 4/x^2) = lim 1/(1 + 4/x + 4/x^2)

lim 1/(1 + 4/x + 4/x^2) = lim 1/(1 + lim4/x + lim4/x^2)

lim 1/(1 + lim4/x + lim4/x^2) = 1/(1+0+0) = 1

Since m = 1 and it is a finite value, we'll calculate n:

n = lim [f(x) - mx] = lim [x^3/(x+2)^2 - x]

lim [x^3/(x+2)^2 - x] = lim (x^3 - x^3 - 4x^2 - 4x)/(x^2 + 4x + 4)

lim (- 4x^2 - 4x)/(x^2 + 4x + 4) = -4/1 = -4

**The equation of the slant asymptote, if x approaches to + infinite and - infinite, is y = x - 4.**