# Math

Graph the function.
g(x)={x+2, x less than or equal to -3
1/2x-2, x>-3

1. 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.

posted by Reiny
2. It's still giving me the wrong answer. and how do I plot -5/2, when the graph shows no decimals?

posted by Kristi
3. 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"

posted by Reiny
4. The table plot on my homework is givving me the wrong answer.

posted by Kristi
5. 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.

posted by Reiny

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