Posted by Kristi on Sunday, October 14, 2012 at 10:09pm.
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.