How do you find the asymptotes for #(3x-2) / (x+1)#?

Answer 1

vertical asymptote at x = - 1
horizontal asymptote at y = 3

The denominator of the function cannot equal zero as this would make the function undefined. Equating the denominator to zero and solving gives the value that x cannot be and if the numerator is non-zero for this value then it is a vertical asymptote.

solve : #x+1=0rArrx=-1" is the asymptote"#

Horizontal asymptotes occur as

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

divide terms on numerator/denominator by x

#f(x)=((3x)/x-2/x)/(x/x+1/x)=(3-2/x)/(1+1/x)#

as #xto+-oo,f(x)to(3-0)/(1+0)#

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

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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.