How do you find the asymptotes for #f(x)=[(5x+3)/(2x-3)]+1#?

1 Answer
Jul 19, 2016

Answer:

vertical asymptote x #=3/2#
horizontal asymptote #y=7/2#

Explanation:

The first step is to express f(x) as a single fraction with common denominator (2x - 3)

#f(x)=(5x+3)/(2x-3)+(2x-3)/(2x-3)=(7x)/(2x-3)#

The denominator of f(x) cannot be zero as this is 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 : #2x-3=0rArrx=3/2" is the asymptote"#

Horizontal asymptotes occur as

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

divide terms on numerator/denominator by x

#((7x)/x)/((2x)/x-3/x)=7/(2-3/x)#

as #xto+-oo,f(x)to7/(2-0)#

#rArry=7/2" is the asymptote"#
graph{(7x)/(2x-3) [-20, 20, -10, 10]}