Question 1194757
Consider the function {{{f(x)=2+1/(x-1)}}}

1. Rewrite this function as a single fraction over the denominator i.

{{{f(x)=2+1/(x-1)}}}

{{{f(x)=(2(x-1))/(x-1)+1/(x-1)}}}

{{{f(x)=(2x-2)/(x-1)+1/(x-1)}}}

{{{f(x)=(2x-2+1)/(x-1)}}}

{{{f(x)=(2x-1)/(x-1)}}}



2. Find the domain of this hyperbola


domain: all {{{x}}} except that makes denominator equal to zero

that is {{{(x-1)=0}}} => {{{x=1}}}

so, domain is: { {{{x}}} element {{{R}}} : {{{x<>1}}} }


3. Explain, in complete sentences, how you use this information to determine the vertical asymptote for this function. Then write the equation of the vertical asymptote.

For rational functions, the vertical asymptotes are the undefined points, also known as the zeros of the denominator, of the simplified function. So, in your case, the vertical asymptote is {{{x=1}}}.


 {{{ drawing( 600, 600, -10, 10, -10, 10, 
green(line(1,-10,1,10)),

graph( 600, 600, -10, 10, -10, 10, (2x-1)/(x-1))) }}}