What is the asymptotes of h(t)=(t^2-2t)/(t^4-16) ?

The asymptotes of h(t)=(t^2-2t)/(t^4-16) is Vertical: t=-2,Horizontal: y=0

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

asymptotes of h(t)=(t^2-2t)/(t^4-16)

asymptotes

h (t) = \frac{t ^{2} - 2 t}{t ^{4} - 16}

Vertical : t = - 2, Horizontal : y = 0

Solution steps

Vertical asymptotes of

\frac{t ^{2} - 2 t}{t ^{4} - 16} : t = - 2

Horizontal Asymptotes of

\frac{t ^{2} - 2 t}{t ^{4} - 16} : y = 0

Slant Asymptotes of

\frac{t ^{2} - 2 t}{t ^{4} - 16} :

None

Vertical : t = - 2, Horizontal : y = 0

f (x) = \frac{1}{2} cot (\frac{x}{2} - \frac{2 π}{3}) - 1

f (x) = 16 x^{2}

x + 6 \cdot (x^{2} + 3)

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

f (x) = - 16 y^{2} + 8 y + 24

What is the asymptotes of h(t)=(t^2-2t)/(t^4-16) ?
The asymptotes of h(t)=(t^2-2t)/(t^4-16) is Vertical: t=-2,Horizontal: y=0