Question 389527
What is the slant asymptote for the function f(x)=(3x^2-4x+3)/(x+2)?

<pre>

{{{"f(x)"}}}{{{""=""}}}{{{(3x^2-4x+3)/(x+2)}}}


Rewrite the function by dividing out the expression on the right
by long division:

     <u>       3x - 10</u>
x + 2)3x² - 4x +  3
      <u>3x² + 6x</u>
          -10x +  3
          <u>-10x - 20</u>
                 23

{{{"f(x)"}}}{{{""=""}}}{{{green(3x-10)}}}{{{red(""+"")}}}{{{red(23/(x+2))}}}
  
That red fraction becomes very small when x is very large in absolute
value.  For instance when x = 1000000, the red fraction {{{red(23/(x+2))}}}
becomes {{{red(23/(x+2))}}} becomes .0000223 which is very near zero.  And also,
when x = -1000000, the red fraction {{{red(23/(x+2))}}} becomes -.000023 which is very near zero.


That means

{{{"f(x)"}}}{{{""=""}}}{{{green(3x-10)}}}{{{red(""+"")}}}{{{red(23/(x+2))}}}

gets closer and closer to just being equal to the green part:

{{{y}}}{{{""=""}}}{{{green(3x-10)}}}

Here is the graph:. The green line is the slant asymptote, whose

equation is

<font color = "green">y = 3x - 10</font>

{{{drawing(400,2000,-10,10,-50,50, graph(400,2000,-10,10,-50,50,(3x^2-4x+3)/(x+2)), green(line(-15,-55, 15,35))  )}}} 

Edwin</pre>