How do you find vertical, horizontal and oblique asymptotes for #f(x) = (3x^2 - 3x- 36) / (2x^2 + 9x +4) #?

1 Answer
May 8, 2016

Answer:

Vertical asymptotes are #x=-1/2# and #x=-4#

and horizontal asymptote #y=3/2#

Explanation:

In #(3x^2-3x-36)/(2x^2+9x+4)#, we get vertical asymptotes by putting denominator as zero i.e. they are given by

#2x^2+9x+4=0# and factorizing by splitting

#2x^2+8x+x+4=0# or #2x(x+4)+1(x+4)=0# or #(2x+1)(x+4)=0#

Hence vertical asymptotes are #x=-1/2# and #x=-4#.

As degree of numerator is equal to that of denominator in the algebraic expression, we have only horizontal asymptote given by

#y=(3x^2)/(2x^2)=3/2#

Had the degree of numerator just one greater than that of denominator, we would have oblique asymptote.

graph{(3x^2-3x-36)/(2x^2+9x+4) [-20, 20, -10, 10]}