What is the asymptotes of g(t)=(t-1)/(t^2+1) ?

The asymptotes of g(t)=(t-1)/(t^2+1) is Horizontal: y=0

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asymptotes of g(t)=(t-1)/(t^2+1)

asymptotes

g (t) = \frac{t - 1}{t ^{2} + 1}

Horizontal : y = 0

Solution steps

Vertical asymptotes of

\frac{t - 1}{t ^{2} + 1} :

None

Horizontal Asymptotes of

\frac{t - 1}{t ^{2} + 1} : y = 0

Slant Asymptotes of

\frac{t - 1}{t ^{2} + 1} :

None

Horizontal : y = 0

y = \frac{2 x ^{2} + x - 18}{x ^{2} - 9}

y + 1 = - \frac{4}{5} = (x - 4)

y = - x

\frac{4 - x ^{2}}{x}

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

What is the asymptotes of g(t)=(t-1)/(t^2+1) ?
The asymptotes of g(t)=(t-1)/(t^2+1) is Horizontal: y=0