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

1 Answer
Oct 25, 2016

Answer:

vertical asymptote at #x=-2/3#
horizontal asymptote at y = 5

Explanation:

The denominator of the rational function cannot be zero as this would make the rational function undefined. Equating the denominator to zero and solving gives the value that x cannot be and if the numerator is non-zero for this value then it is a vertical asymptote.

solve: #3x+2=0rArrx=-2/3" is the asymptote"#

Horizontal asymptotes occur as

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

divide terms on numerator/denominator by x

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

as #xto+-oo,f(x)to(9-0)/(3+0)+2#

#rArry=5" 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 1 ) Hence there are no slant asymptotes.
graph{(9x-4)/(3x+2)+2 [-20, 20, -10, 10]}