Question 868612
<pre>
We draw the graph of y=4/x, by getting and plotting some points:

 x|y
--|--
±1|±4
±2|±2
±3|±1.3
±4|±1

{{{drawing(300,300,-10,10,-10,10,

graph(300,300,-10,10,-10,10,4/x),
circle(1,4,0.15),circle(1,4,0.13),circle(1,4,0.11),circle(1,4,0.09),
circle(1,4,0.07),circle(1,4,0.05),circle(1,4,0.03),circle(1,4,0.01),

circle(-1,-4,0.15),circle(-1,-4,0.13),circle(-1,-4,0.11),circle(-1,-4,0.09),
circle(-1,-4,0.07),circle(-1,-4,0.05),circle(-1,-4,0.03),circle(-1,-4,0.01),


circle(2,2,0.15),circle(2,2,0.13),circle(2,2,0.11),circle(2,2,0.09),
circle(2,2,0.07),circle(2,2,0.05),circle(2,2,0.03),circle(2,2,0.01),

circle(-2,-2,0.15),circle(-2,-2,0.13),circle(-2,-2,0.11),circle(-2,-2,0.09),
circle(-2,-2,0.07),circle(-2,-2,0.05),circle(-2,-2,0.03),circle(-2,-2,0.01),

circle(3,1.33333333,0.15),circle(3,1.33333333,0.13),circle(3,1.33333333,0.11),
circle(3,1.33333333,0.09),circle(3,1.33333333,0.07),circle(3,1.33333333,0.05),
circle(3,1.33333333,0.03),circle(3,1.33333333,0.01),

circle(-3,-1.33333333,0.15),circle(-3,-1.33333333,.13),
circle(-3,-1.33333333,0.11),circle(-3,-1.33333333,.09),
circle(-3,-1.33333333,0.07),circle(-3,-1.33333333,.05),
circle(-3,-1.33333333,0.03),circle(-3,-1.33333333,.01),

circle(4,1,0.15),circle(4,1,0.13),circle(4,1,0.11),
circle(4,1,0.09),circle(4,1,0.07),circle(4,1,0.05),
circle(4,1,0.03),circle(4,1,0.01),

circle(-4,-1,0.15),circle(-4,-1,0.13),
circle(-4,-1,0.11),circle(-4,-1,0.09),
circle(-4,-1,0.07),circle(-4,-1,0.05),
circle(-4,-1,0.03),circle(-4,-1,0.01) )}}}

Now the lines y=mx+4 will all have y intercept (0,4),
but will have different slopes.  Here are some lines
that have the equation y=mx+4 where m has different
values


{{{drawing(300,300,-10,10,-10,10,

graph(300,300,-10,10,-10,10,4/x,-x+4,15,4,-3x+4,2x+4,-(104/225)x+4)  )}}}

Some of those lines intersect the graph in 0 point, 1 points or 2 points.

The brownish-green line does not intersect it at all
The green and horizontal intersect the graph in 1 point
The purple and the light blue lines intersect the curve
in two points each.

If we solve the system of equations:

{{{system(y=mx+4,y=4/x)}}}

by substitution, we get

{{{mx+4}}}{{{""=""}}}{{{4/x}}}

{{{mx^2+4x}}}{{{""=""}}}{{{4}}}

{{{mx^2+4x-4}}}{{{""=""}}}{{{"0"}}}

What determines whether the line y=mx+4 intersects
the curve 0, 1, or 2 times, is whether this
quadratic equation has 0,1, or 2 real solutions.

And what determines whether a quadratic equation has
0, 1, or 2 real solutions is the DISCRIMINANT {{{b^2-4ac}}}.

{{{mx^2+4x-4}}}{{{""=""}}}{{{"0"}}} 

a=m, b=4, c=-4 so the 

discriminant = {{{b^2-4ac}}} = {{{4^2-4m(-4)}}} = {{{16+16m}}}

Case 1:  If the discriminant is negative, there will be NO real solutions.

So we set the discriminant less than 0

{{{16+16m<0}}}
      {{{16m<-16}}}
       {{{m<-4}}}

So when m is less than -4, there will be NO solutions.

So we set the discriminant less than 0

{{{16+16m<0}}}
      {{{16m<-16}}}
       {{{m<-1}}}

So when m is less than -1, there will be NO solutions.
and therefore the line will NOT intersect the curve at all.


Case 2: If the discriminant is EQUAL to zero, there will be ONE 
real solution, and therefore the line will intersect the curve ONE time.

So we set the discriminant EQUAL to 0

{{{16+16m=0}}}
      {{{16m=-16}}}
       {{{m=-1}}}

So when m is equal to -1, there will be ONE real solutions.
and therefore the line will intersect the curve ONE time.

Case 3: If the discriminant is positive, there will be TWO real solutions.

So we set the discriminant GREATER than 0

{{{16+16m>0}}}
      {{{16m>-16}}}
       {{{m>-1}}}

So when m is greater than -1, there will be TWO solutions,
and therefore the line will intersect the curve at TWICE.

Answer:

If the slope m < -1, the line y=mx+4 will intersect the curve 0 times.
If the slope m = -1, the line y=mx+4 will intersect the curve 1 time.
If the slope m > -1, the line y=mx+4 will intersect the curve 2 times.
 
Edwin</pre>