Question 685713
<pre>
T(w)={{{(w^2+2w+4)/(w^2+1)}}}

Let's change w to x and T(w) to y

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

There are no vertical asymptotes because when
we set the denominator = 0 

     x²+1 = 0
       x² = -1
there are no real solutions to that so there are 
no vertical asymptotes.

There is a horizontal asymptote because numerator
and denominator are both of the same degree 2 and therefore
the horizontal asymptote is y = the ratio of the leading
coefficients of numerator and denominator which is y = {{{1/1}}}
or y = 1

The y-intercept is found by substituting 0 for x

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

So the y-intercept is (0,4)

There are no real zeros because when we set y=0

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

We multiply both sides by the denominator and have:

x²+2x+4 = 0

The discriminant b²-4ac = 2²-4·1·4 = 4-16 = -12
so there are no real solutions to that.  So there
are no x-intercepts   Since the y-intercept is
above the x-axis, and since there are no vertical
asymptotes the entire curve is above the x-axis.

Let's see if it crosses its horizontal asymptote 
y=1

y ={{{(x^2+2x+4)/(x^2+1)}}}
1 ={{{(x^2+2x+4)/(x^2+1)}}}
 
x² + 1 = x² + 2x + 4
    -3 = 2x
   {{{-3/2}}} = x

So it crosses its horizontal asymptote at ({{{-3/2}}},1)  

Let's draw what we have the y-intercept (0,4), the
horizontal asymptote y = 1 (in green) and the point ({{{-3/2}}},1),
where the curve crosses its horizontal asymptote:

{{{drawing(400,200,-8,8,-2.5,5.5,

graph(400,200,-8,8,-2.5,5.5), circle(0,4,.1),circle(-3/2,1,.1),
green(line(-20,1,20,1))  )}}} 
All we can do now is get some points:

 x| y
------
-8|.80
-6|.76
-4|.70
-2|.80
 1|3.5
 3|1.9
 7|1.2

{{{drawing(400,200,-8,8,-2.5,5.5,

graph(400,200,-8,8,-2.5,5.5), circle(0,4,.1),circle(-3/2,1,.1),
green(line(-20,1,20,1)),
circle(-4,((-4)^2+2(-4)+4)/((-4)^2+1),.08),
circle(-2,((-2)^2+2(-2)+4)/((-2)^2+1),.08),
circle(1,((1)^2+2(1)+4)/((1)^2+1),.08),
circle(3,((3)^2+2(3)+4)/((3)^2+1),.08),
circle(7,((7)^2+2(7)+4)/((7)^2+1),.08),
circle(-8,((-8)^2+2(-8)+4)/((-8)^2+1),.08), 
circle(-6,((-6)^2+2(-6)+4)/((-6)^2+1),.08))}}}and sketch the curve:{{{drawing(400,200,-8,8,-2.5,5.5,

graph(400,200,-8,8,-2.5,5.5,(x^2+2x+4)/(x^2+1)), circle(0,4,.1),circle(-3/2,1,.1),
green(line(-20,1,20,1)),
circle(-4,((-4)^2+2(-4)+4)/((-4)^2+1),.08),
circle(-2,((-2)^2+2(-2)+4)/((-2)^2+1),.08),
circle(1,((1)^2+2(1)+4)/((1)^2+1),.08),
circle(3,((3)^2+2(3)+4)/((3)^2+1),.08),
circle(7,((7)^2+2(7)+4)/((7)^2+1),.08),
circle(-8,((-8)^2+2(-8)+4)/((-8)^2+1),.08), 
circle(-6,((-6)^2+2(-6)+4)/((-6)^2+1),.08))}}}

Edwin</pre>