Question 65647
<pre>we have graphed the boundary line for the linear inequality. Determine the
correct half-plane in each case, and complete the graph. 
Y > 3
THE GRAPH THAT IS ON THE BOOK ITS A LINE JUST GOING THROUGH ALL 3 ON THE TOP
TWO THE NEGATIVE SIDE AND POSITIVE SIDE LIKE THIS: 
<----------3------------------> 
i DON'T KNOW HOW TO ANSWER THIS AND I DON'T EVEN UNDERSTAND WHAT IS ASKING
<font size = 4><b>

A "half-plane" is all the area on just one side of a line.

The boundary line for the INEQUALITY 

y > 3

has the EQUATION

y = 3


To draw the graph of y = 3, plot some points.  Since there is only
one letter in y = 3, we must choose the y-value of every point to
be 3, but since there is no x, you can choose x to be any number.
I'll arbitrarily pick 3, -2, 5, and -4 for x, and only 3 for y.

 x|y
-----
 3|3
-2|3
 5|3
-4|3

So plot these points (3,3), (-3,3), (5,3), and (-4,3)

They all line on this horizontal line:

{{{ graph( 300, 300, -7, 7, -7, 7, 3)}}}

That horizontal line should be drawn dotted, not solid
as I have here, but I can only draw solid lines on here,
but you can draw it dotted on your paper.  (If the
inequality symbol is > or <, then you draw the boundary
line dotted.  If the inequality symbol is <u>></u> or
<u><</u>, you draw the boundary line solid.

If the inequality starts off " y > " or " y <u>></u> " 
you shade the area ABOVE the line. 

If the inequality starts off " y < " or " y <u><</u> "
you shade the area BELOW the line.

Your inequality starts off " y > ", so you shade the
area ABOVE the line, like where these slanted lines are

{{{ graph( 300, 300, -7, 7, -7, 7, 3, (x-3)*(sqrt(-6+x))/(sqrt(-6+x)), (x+12)*(sqrt(9+x))/(sqrt(9+x)), (x+10)*(sqrt(7+x))/(sqrt(7+x)), (x+8)*(sqrt(5+x))/(sqrt(5+x)), (x+6)*(sqrt(3+x))/(sqrt(3+x)),(x+4)*(sqrt(1+x))/(sqrt(1+x)),(x+2)*(sqrt(-1+x))/(sqrt(-1+x)) ) }}}

But don't forget to draw the horizontal line dotted instead of
solid as I have it here, and to shade the area ABOVE the line.

The area you shade is called a "half plane"

Edwin</pre>