What are the asymptotes for #y=3/(x-1)+2# and how do you graph the function?

We are given the rational function #color(green)( f(x) = [3/(x-1)] + 2#

We will simplify and rewrite #f(x)# as

#rArr [3+2(x-1)]/(x-1)#

#rArr [3+2x-2]/(x-1)#

#rArr [2x+1]/(x-1)#

Hence,

# color(red)(f(x) = [2x+1]/(x-1))#

Vertical Asymptote

Set the denominator to Zero.

So, we get

#(x-1) = 0#

#rArr x = 1#

Hence,

Vertical Asymptote is at #color(blue)(x = 1#

Horizontal Asymptote

We must compare the degrees of the numerator and denominator and verify whether they are equal.

To compare, we need to deal with **lead coefficients. **

The lead coefficient of a function is the number in front of the term with the highest exponent.

If our function has a horizontal asymptote at # color(red)(y = a / b)#,

where #color(blue)(a)# is the lead coefficient of the numerator, and

#color(blue)b# is the lead coefficient of the denominator.

#color(green)(rArr y = 2/1)#

#color(green)(rArr y = 2)#

Hence,

Horizontal Asymptote is at #color(blue)(y = 2#

Graph of the rational function with the horizontal asymptote and the vertical asymptote can be found below:

I hope you find this solution with the graph useful.