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.