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

1 Answer
Aug 5, 2016

Answer:

vertical asymptotes #x=-4,x=-1/2#
horizontal asymptote #y=3/2#

Explanation:

The denominator of f(x) cannot be zero as this is undefined. Equating the denominator to zero and solving gives the values that x cannot be and if the numerator is non-zero for these values then they are asymptotes.

solve: #2x^2+9x+4)=0rArr(2x+1)(x+4)=0#

#rArrx=-4" and " x=-1/2" are the asymptotes"#

Horizontal asymptotes occur as

#lim_(xto+-oo),f(x)toc" (a constant)"#

divide terms on numerator/denominator by the highest power of x, that is #x^2#

#((3x^2)/x^2-(3x)/x^2-36/x^2)/((2x^2)/x^2+(9x)/x^2+4/x^2)=(3-3/x-36/x^2)/(2+9/x+4/x^2)#

as #xto+-oo,f(x)to(3-0-0)/(2+0+0)#

#rArry=3/2" is the asymptote"#

Slant asymptotes occur when the degree of the numerator > degree of the denominator. This is not the case here (both of degree 2 ) Hence there are no slant asymptotes.
graph{(3x^2-3x-36)/(2x^2+9x+4) [-20, 20, -10, 10]}