What is the asymptotes of f(x)=(-4x^2-3x+8)/(x+1) ?

The asymptotes of f(x)=(-4x^2-3x+8)/(x+1) is Vertical: x=-1,Slant: y=-4x+1

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asymptotes of f(x)=(-4x^2-3x+8)/(x+1)

asymptotes

f (x) = \frac{- 4 x ^{2} - 3 x + 8}{x + 1}

Vertical : x = - 1, Slant : y = - 4 x + 1

Solution steps

Vertical asymptotes of

\frac{- 4 x ^{2} - 3 x + 8}{x + 1} : x = - 1

Horizontal Asymptotes of

\frac{- 4 x ^{2} - 3 x + 8}{x + 1} :

None

Slant Asymptotes of

\frac{- 4 x ^{2} - 3 x + 8}{x + 1} : y = - 4 x + 1

Vertical : x = - 1, Slant : y = - 4 x + 1

x - 1, x < 0

f (x) = 3 x^{4} - 4 x^{3}

f (x) = (x + 4)^{\frac{4}{7}}

f (x) = lo g_{3} (9 x)

\frac{2 x ^{2} - 3}{x ^{2} + 2 x + 1}

What is the asymptotes of f(x)=(-4x^2-3x+8)/(x+1) ?
The asymptotes of f(x)=(-4x^2-3x+8)/(x+1) is Vertical: x=-1,Slant: y=-4x+1