How do you find the asymptotes for #f(x) = (3x^2 + 15x +18) /( 4x^2-4)#?

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Answer 1 · 2017-01-03T08:08:41

The vertical asymptotes are #x=1# and #x=-1#
No slant asymptote
The horizontal asymptote is #y=3/4#

We need

#a^2-b^2=(a+b)(a-b)#

We factorise the denominator

#4x^2-4=4(x^2-1)=4(x-1)(x+1)#

The domain of #f(x)# is #D_f(x)=RR-{-1,1}#

As we cannot divide by #0#, #x!=-1# and #x!=1#

The vertical asymptotes are #x=1# and #x=-1#

As the degree of the numerator is #=# to the degree of the denominator, there is no slant asymptote

To find the horizontal asymptotes, we look at the limits of #f(x)# as #x->+-oo#

#lim _(x->+-oo) f(x )#

#= lim _(x->+-oo) (3x^2) / (4x^2)#

#=3/4#

The horizontal asymptote is #y=3/4#

graph{(y-(3x^2+15x+18)/(4x^2-4))(y-3/4)=0 [-14.24, 14.23, -7.12, 7.11]}