Math
posted by Kristi .
Graph the function.
g(x)={x+2, x less than or equal to 3
1/2x2, x>3

for x < 3 , sketch the line y = x+2
put an "open" point at (3,1) , showing the point is excluded
for x≥3, sketch y = (1/2)x  2
draw a "closed' point at (3, 5/2) , showing that point is included. 
It's still giving me the wrong answer. and how do I plot 5/2, when the graph shows no decimals?

If you are doing this kind of math, I am surprised that you didn't know that
5/2 = 2.5 or 2 1/2
What do you mean by "it's still giving me the wrong answer" , what is "it" 
The table plot on my homework is givving me the wrong answer.

I did not read your question carefully enough
change it to:
for x ≤ 3 , sketch the line y = x+2
put an "closed" point at (3,1) , showing the point is included
for x>3, sketch y = (1/2)x  2
draw a "open' point at (3, 7/2) , showing that point is excluded.
notice the second point is (3, 7/2) or (3, =3.5)
try it now.
The only other thing that could be wrong is that your second equation is
y = 1/(2x) 2 , but I doubt it.