What is the asymptotes of f(x)=(6x^2+1)/(2x^2-3) ?

The asymptotes of f(x)=(6x^2+1)/(2x^2-3) is Vertical: x=-sqrt(3/2),x=sqrt(3/2),Horizontal: y=3

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asymptotes of f(x)=(6x^2+1)/(2x^2-3)

asymptotes

f (x) = \frac{6 x ^{2} + 1}{2 x ^{2} - 3}

Vertical : x = - \frac{3}{2}, x = \frac{3}{2}, Horizontal : y = 3

Solution steps

Vertical asymptotes of

\frac{6 x ^{2} + 1}{2 x ^{2} - 3} : x = - \frac{3}{2}, x = \frac{3}{2}

Horizontal Asymptotes of

\frac{6 x ^{2} + 1}{2 x ^{2} - 3} : y = 3

Slant Asymptotes of

\frac{6 x ^{2} + 1}{2 x ^{2} - 3} :

None

Vertical : x = - \frac{3}{2}, x = \frac{3}{2}, Horizontal : y = 3

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

f (x) = x - 5

f (x) = - x^{2} - 9

x + 2

f (x) = 2 x^{\frac{1}{5}} - 4

What is the asymptotes of f(x)=(6x^2+1)/(2x^2-3) ?
The asymptotes of f(x)=(6x^2+1)/(2x^2-3) is Vertical: x=-sqrt(3/2),x=sqrt(3/2),Horizontal: y=3