Question 1157352
{{{R(x)=(2x^2-3x+2)/(x-1)}}}
<pre>
The vertical asymptote is found by setting the denominator = 0.

x-1 = 0
  x = 1    <-- equation of the vertical asymptote:

Draw the vertical asymptote (in green):

{{{drawing(400/2,(23600/29)/2,-2.9,2.9,-4.9,6.9,
green(line(1,-10,1,10)),
graph(400/2,(23600/29)/2,-2.9,2.9,-4.9,6.9) )}}}

A rational function has an oblique asymptote when and only when
the degree of the numerator is exactly 1 more than the degree of 
the denominator.  That is the case here, so

we use long division

   <u>     2x-1</u>
x-1)2x²-3x+2
    <u>2x²-2x</u>
        -x+2 
        <u>-x+1</u>
           1

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

The oblique asymptote of a rational function has the equation 
y = the quotient
when the rational function's equation is divided out.  The
remainder must not be 0 and is ignored.

Therefore the oblique asymptote here has the equation y = 2x-1

{{{drawing(400/2,(23600/29)/2,-2.9,2.9,-4.9,6.9,
green(line(1,-10,1,10),line(-10,-21,10,19)),
graph(400/2,(23600/29)/2,-2.9,2.9,-4.9,6.9) )}}}

Then we sketch in the graph:

{{{drawing(400/2,(23600/29)/2,-2.9,2.9,-4.9,6.9,
green(line(1,-10,1,10),line(-10,-21,10,19)),
graph(400/2,(23600/29)/2,-2.9,2.9,-4.9,6.9,(2x^2-3x+2)/(x-1)) )}}}

Edwin</pre>