How do I find the asymptotes of #y=-1/x^2#?

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Answer 1 · 2015-09-13T20:28:34

Horizontal asymptote at: #y = 0#

Vertical asymptote at: #x= 0#

#y=-1/x^2#

The horizontal asymptote is at : #lim(x to +-oo#) #(y)#

So in this case:

#lim(x to +-oo#) #(-1/x^2) =>#

as x tends to infinity the denominator keeps getting larger and

larger forcing the fraction to tend to zero, hence: the horizontal

asymptote is at #y = 0#, AKA the x-axis

The vertical asymptote(s) are the at the value(s) of x that make the

function #f(x) = y# undefined hence the vertical asymptote in this

case is at:

#x = 0#, AKA the y-axis.