Question

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

Answer 1

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]}