How do you find the vertical, horizontal or slant asymptotes for #(4x^2+11) /( x^2+8x−9)#?

Question

James Allen · Accepted Answer

The vertical asymptotes are #x=-9# and #x=1#
No slant asymptote.
The horizontal asymptote is #y=4# Let's factorise the denominator #x^2+8x-9=(x-1)(x+9)# Let, #f(x)=(4x^2+11)/(x^2+8x-9)# The domain of #f(x)# is #D_f(x)=RR-{-9,1} # As we cannot divide by #0#, #x!=-9# and #x!=1# So, the vertical asymptotes are #x=-9# and #x=1# The degree of the numerator is #=# to the degree of the denominator, we don't have a slant asymptote. To find the horizontal asymptote, we compute the limits of #f(x)# as #x->+-oo# #lim_(x->+-oo)f(x)=lim_(x-+-oo)4x^2/x^2=4# The horizontal asymptote is #y=4# graph{(y-(4x^2+11)/(x^2+8x-9))(y-4)(x-1)(x+9)=0 [-52, 52, -26, 26]}

Theodore Amidon · Answer

undefined To find the vertical asymptotes of a rational function, set the denominator equal to zero and solve for $ x $. These values of $ x $ correspond to vertical asymptotes. In this case, the denominator $ x^2 + 8x - 9 $ factors as $ (x + 9)(x - 1) = 0 $, so the vertical asymptotes occur at $ x = -9 $ and $ x = 1 $.

To find horizontal asymptotes, compare the degrees of the numerator and denominator. If the degree of the numerator is less than the degree of the denominator, the horizontal asymptote is at $ y = 0 $. If the degrees are equal, divide the leading coefficient of the numerator by the leading coefficient of the denominator to find the horizontal asymptote. If the degree of the numerator is greater, there is no horizontal asymptote.

In this case, since the degree of the numerator (2) is equal to the degree of the denominator (2), divide the leading coefficient of the numerator (4) by the leading coefficient of the denominator (1) to find the horizontal asymptote, which is $ y = 4 $.

There are no slant asymptotes for this rational function since the degree of the numerator is not greater than the degree of the denominator by exactly one.