Question 905910
<pre>
You are a bit confused, so let's see if we can straighten you out.

{{{drawing(400,400,-5,5,-3,5,graph(400,400,-5,5,-3,5,(x^2+6x+7)*sqrt(-x-.9)/sqrt(-x-.9)),

graph(400,400,-5,5,-3,5,(-2x^2+4)*sqrt(1-abs(x-.03))/sqrt(1-abs(x-.03))),

graph(400,400,-5,5,-3,5,(x^2-4x+5)*sqrt(x-1)/sqrt(x-1)),

green(line(-3,10,-3,-10),line(2,10,2,-10)),

locate(-3,-2,"(-3,-2)"),locate(.05,4.2,"(0,4)"),locate(2,1,"(2,1)"),
locate(-5,2.4,matrix(1,2,from,-infinity)),locate(4,5,matrix(1,2,to,""+infinity))




 )}}}

<pre>
It is unfortunate that the people who made mathematics gave open 
intervals and points the same kind of notation.  A point has an
x-value and a y-value, but an interval has two x-values.

So be EXTREMELY careful to notice the difference between the point
(a,b) and the interval (a,b), in the following:

From negative infinity to the point (-3,-2) the graph is falling, so it
is decreasing from x = negative infinity to x = -3.

So the graph is DECREASING on the interval (-oo,-3).  

From the point (-3,-2) to the point (0,4) the graph is rising, so it
is increasing from x=-3 to x=0

So the graph is INCREASING on the interval (-2,0).

From the point (0,4) to the point (2,1) the graph is falling again,
so it is decreasing again from x=0 to x=2.

So the graph is DECREASING on the interval (0,2).

From the point (2,1) to positive infinity the graph is rising,
so it is increasing from x=2 to x=oo

So the graph is INCREASING on the interval (2,oo)

-----
Answer: increasing on the intervals (-3,0) and (2,oo).
        decreasing on the intervals (-oo,-3) and (0,2).

Edwin</pre>