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What are the asymptotes of #f(x)=-x/((2x-3)(7x-1)) #?

Answer 1

vertical asymptotes at #x=1/7" and "x=3/2#
horizontal asymptote at y = 0

The denominator of f(x) cannot be zero as this would make f(x) undefined. Equating the denominator to zero and solving gives the values that x cannot be and if the numerator is non-zero for these values then they are vertical asymptotes.

solve : #(2x-3)(7x-1)=0#

#rArrx=1/7" and "x=3/2" are the asymptotes"#

Horizontal asymptotes occur as

#lim_(xto+-oo),f(x)toc" (a constant)"#

divide terms on numerator/denominator by the highest power of x, that is #x^2#

#f(x)=-x/(14x^2-23x+3)=-(x/x^2)/((14x^2)/x^2-(23x)/x^2+3/x^2)#

#=-(1/x)/(14-23/x+3/x^2)#

as #xto+-oo,f(x)to0/(14-0+0)#

#rArry=0" is the asymptote"# graph{-(x)/((2x-3)(x-1)) [-20, 20, -10, 10]}

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Answer 2

Charles Arocha

The asymptotes of f(x)=-x/((2x-3)(7x-1)) are vertical asymptotes at x = 3/2 and x = 1/7, and there is no horizontal asymptote.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.