How do you find the vertical, horizontal or slant asymptotes for #(x + 2)/( x + 3)#?

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Answer 1 · 2016-05-28T07:54:53

#(x+2)/(x+3)# has a vertical asymptote #x=-3# and a horizontal asymptote #y=1#

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

When #x=-3#, the denominator is zero and the numerator non-zero, so #f(-3)# is undefined and #f(x)# has a vertical asymptote there.

Note that:

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

As #x->+-oo#, the expression #-1/(x+3)->0#

So #f(x)->1+0 = 1# as #x->+-oo# and #f(x)# has a horizontal asymptote #y = 1#.

graph{(x+2)/(x+3) [-13.21, 6.79, -4.24, 5.76]}