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

Question

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

Christian Antunez · Accepted Answer

#(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]}

Lily Anderson · Answer

undefined To find the vertical asymptote(s), set the denominator equal to zero and solve for $ x $. In this case, the denominator is $ x + 3 $, so:

$ x + 3 = 0 $

$ x = -3 $

Therefore, the vertical asymptote is $ x = -3 $.

There are no horizontal asymptotes for rational functions of the form $ \frac{f(x)}{g(x)} $, where $ f(x) $ and $ g(x) $ are polynomials, if the degree of the numerator is less than the degree of the denominator. Since the degree of the numerator is 1 and the degree of the denominator is also 1, there are no horizontal asymptotes.

To find slant asymptotes, divide the numerator by the denominator using polynomial long division. In this case:

$ \frac{x + 2}{x + 3} = 1 - \frac{1}{x + 3} $

Since the degree of the numerator is less than the degree of the denominator after division, there are no slant asymptotes.