Question 687897
<pre>
T(w)= {{{(w^2+2w+4)/(w-1)}}} 

Change Y(w) to y and w to x

y = {{{(x^2+2x+4)/(x-1)}}}

That has a vertical asymptote.  Set denominator = 0

x-1 = 0
  x = 1

So here is the vertical asymptote:
{{{drawing(400,720, -10,10,-16,20,graph(400,720, -10,10,-16,20),
green(line(1,50,1,-50))
 )}}}

It has no horizontal asymptote because the numerator's largest
power of x (its degree) which is 2 is larger that the denominator's largest
power of x (its degree) with is 1.  

However since the degree of the numerator is exactly one more
than the degree of the denominator, it has a slanted or "oblique" asymptote.

We find that by long division:

    <u>    x+3</u>
x-1)x²+2x+4
    <u>x²- x</u>
       3x+4
       <u>3x-3</u>
          7

So y = x+3 + {{{7/(x-1)}}}

And the fraction {{{7/(x-1)}}} gets smaller and smaller as x gets
larger and larger either positively or negatively, so the curve
approaches just being like the line y = x+3 ignoring the remainder
or denominator.  So we draw the slant asymptote y = x+3

   {{{drawing(400,720,-10,10,-16,20,graph(400,720, -10,10,-16,20),green(line(1,50,1,-50), line(-20,-17,13,16))  )}}}

The y-intercept is gotten by substituting x=0

y = {{{(x^2+2x+4)/(x-1)}}}
y = {{{(0^2+2*0+4)/(0-1)}}}
y = {{{4/(-1)}}}
y = -4

So the y-intercept is (0,-4). There is no x-intercept because the numerator
is never zero. So we plot the y-intercept and get a few more points, maybe
(8,12),(-6,-4), (2,12), (-4,-2.4), (3,9.5), (-1,-1.5). and draw the graph:
{{{drawing(400,720,-10,10,-16,20,graph(400,720, -10,10,-16,20,(x^2+2x+4)/(x-1)),green(line(1,50,1,-50), line(-20,-17,13,16))  )}}}

Edwin</pre>