SOLUTION: Graph the following rational function showing all work, giving zeros, asymptotes, and y-intercept. Example question(s): T(w)={{{(w^2+2w+4)/(w^2+1)}}}{

T(w)=

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

y =

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 = 
or y = 1

The y-intercept is found by substituting 0 for x

y =
y = 4

So the y-intercept is (0,4)

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

0 = 

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 =
1 =
 
x² + 1 = x² + 2x + 4
    -3 = 2x
    = x

So it crosses its horizontal asymptote at (,1)  

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

 
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

and sketch the curve:

Edwin