What is the asymptotes of f(x)=(x^3+1)/(x^2+4x) ?

The asymptotes of f(x)=(x^3+1)/(x^2+4x) is Vertical: x=-4,x=0,Slant: y=x-4

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asymptotes of f(x)=(x^3+1)/(x^2+4x)

asymptotes

f (x) = \frac{x ^{3} + 1}{x ^{2} + 4 x}

Vertical : x = - 4, x = 0, Slant : y = x - 4

Solution steps

Vertical asymptotes of

\frac{x ^{3} + 1}{x ^{2} + 4 x} : x = - 4, x = 0

Horizontal Asymptotes of

\frac{x ^{3} + 1}{x ^{2} + 4 x} :

None

Slant Asymptotes of

\frac{x ^{3} + 1}{x ^{2} + 4 x} : y = x - 4

Vertical : x = - 4, x = 0, Slant : y = x - 4

y = (2 x - 8)^{\frac{2}{3}}

f (x) = - \frac{1}{x + 8}

(x - 3)^{2} + 2

f (x) = \frac{x ^{2}}{x - 4}

f (x) = \frac{( 1 - 5 t )}{6 + t}

What is the asymptotes of f(x)=(x^3+1)/(x^2+4x) ?
The asymptotes of f(x)=(x^3+1)/(x^2+4x) is Vertical: x=-4,x=0,Slant: y=x-4