SOLUTION: can someone please help me with this problem.....Find the horizontal or oblique asymptote if any of R(x)=(2x^2-3x+2)/(x-1)____ please explain in detail

Algebra ->  Rational-functions -> SOLUTION: can someone please help me with this problem.....Find the horizontal or oblique asymptote if any of R(x)=(2x^2-3x+2)/(x-1)____ please explain in detail      Log On


   

Question 1157352: can someone please help me with this problem.....Find the horizontal or oblique asymptote if any of R(x)=(2x^2-3x+2)/(x-1)____ please explain in detail
Answer by Edwin McCravy(20064) About Me  (Show Source):
You can put this solution on YOUR website!
R%28x%29=%282x%5E2-3x%2B2%29%2F%28x-1%29
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):



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

        2x-1
x-1)2x²-3x+2
    2x²-2x
        -x+2 
        -x+1
           1

R%28x%29=2x-1%2B1%2F%28x-1%29

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



Then we sketch in the graph:



Edwin