Question 1100446
<pre>
{{{y}}}{{{""=""}}}{{{(x^2+3x-4)/(x^2-x-20)}}}

Factor the top and bottom:

{{{y}}}{{{""=""}}}{{{((x+4)(x-1))/((x-5)(x+4))}}}

Now as long as you do not cancel the (x+4)'s, then
x cannot equal to -4, because if you substitute
x = -4, you get:

{{{y}}}{{{""=""}}}{{{((-4+4)(-4-1))/((-4-5)(-4+4))}}}

{{{y}}}{{{""=""}}}{{{((0)(-5))/((-9)(0))}}}

{{{y}}}{{{""=""}}}{{{0/0}}}

That is undefined, because we cannot divide anything
by 0, not even 0 divided by 0.  It's simply undefined.
That's what causes the hole to be in the graph.

Now if we cancel the (x+4)'s we get a graph which has
no hole.

{{{y}}}{{{""=""}}}{{{((cross(x+4))(x-1))/((x-5)(cross(x+4)))}}}

{{{y}}}{{{""=""}}}{{{(x-1)/(x-5)}}}

Its graph does NOT have a hole!

Notice that if we substitute x=-4 in it, we get:

{{{y}}}{{{""=""}}}{{{(-4-1)/(-4-5)}}}{{{""=""}}}{{{5/9}}}

So the graph goes through the point {{{(matrix(1,3,-4,",",5/9))}}}

So the graph of

{{{y}}}{{{""=""}}}{{{(x-1)/(x-5)}}}

whih has no hole, is this:

{{{drawing(400,400,-6,10,-7,9,graph(400,400,-6,10,-7,9,(x-1)/(x-5)),
circle(-4,5/9,.2),circle(-4,5/9,.08),circle(-4,5/9,.06),circle(-4,5/9,.04),circle(-4,5/9,.02),circle(-4,5/9,.01),circle(-4,5/9,.008),circle(-4,5/9,.09),circle(-4,5/9,.1),locate(-5.8,2.1,(matrix(1,3,-4,",",5/9)))

 )}}}

That's the graph WITHOUT the hole because the point {{{(matrix(1,3,-4,",",5/9))}}} is there, indicated by 
the darkened circle.

Now let's go back to the original equation, which has a hole,
because the (x+4)'s were not canceled:

{{{y}}}{{{""=""}}}{{{(x^2+3x-4)/(x^2-x-20)}}}{{{""=""}}}{{{((x+4)(x-1))/((x-5)(x+4))}}}

Its graph is the same as the graph of {{{y}}}{{{""=""}}}{{{(x-1)/(x-5)}}} except
that it has a hole and doesn't go through the point {{{(matrix(1,3,-4,",",5/9))}}}. Its graph is:

{{{drawing(400,400,-6,10,-7,9,

graph(400,400,-6,10,-7,9,((x-1)/(x-5))*sqrt(-4.2-x)/sqrt(-4.2-x)),

graph(400,400,-6,10,-7,9,((x-1)/(x-5))*sqrt(3.8+x)/sqrt(3.8+x)),


circle(-4,5/9,.2),locate(-5.8,2.1,(matrix(1,3,-4,",",5/9)))

 )}}}

So when you leave the equation as it was given originally,

{{{y}}}{{{""=""}}}{{{(x^2+3x-4)/(x^2-x-20)}}}

And do not factor and cancel out the (x+4)'s then you have a
hole at the point that makes (x+4) equal to 0. 

Then when you factor and cancel the (x+4)'s, you "fill in" the 
hole.  There is a hole when you don't cancel, and no hole when
you do cancel.

Edwin</pre>