Graph the function (piecewise)

[ghostly]

Graph the Function.

f(x)={x-2, for x>0
{-5, for x is less than or equal to 0

I'm lost lol. Any help would be great thanks.

I don't know if you plug in a zero for f(x)={-5, for x is less than or equal to 0 to make it (-5,0) and then plug in a 1 to make it (-5,1) making it a vertical line with a closed point on (-5,0)

And Also I dont know if its an open point a (0,-2) with points (1,-1), (2,0) etc.

 

[ghostly]

I think I may have found the answer....I don't know how to insert a graph so ill try to explain. I have an open circle at (0,-2) with a slope of 1 over 1 for f(x)= {x-2, for x>0(greater goes to the right). I have a closed point at (0,-5) with a slope of 0(horizontal line) going left because its {-5, for x is less than or equal to zero. Can anyone confirm?

 

[HallsofIvy]

Graph the Function.

f(x)={x-2, for x>0
{-5, for x is less than or equal to 0

I'm lost lol. Any help would be great thanks.

I don't know if you plug in a zero for f(x)={-5, for x is less than or equal to 0 to make it (-5,0) and then plug in a 1 to make it (-5,1) making it a vertical line with a closed point on (-5,0)

The line from (-5, 1) to (-5, 0) is horizontal, not vertical.

And Also I dont know if its an open point a (0,-2) with points (1,-1), (2,0) etc.

y= x- 2 is a straight line starting at (0, -2) and going up to the right with slope 1. y= -5 is a horizontal line. Since the problem says "x- 2, for x> 0", x=0 does NOT belong to this part of the grapy. Use an "open point" to show that. Since it says "-5, for x less than or equal to 0", x= 0 is included in this part of the graph. Use a "closed point" to show that.

 

[ghostly]

The line from (-5, 1) to (-5, 0) is horizontal, not vertical.


y= x- 2 is a straight line starting at (0, -2) and going up to the right with slope 1. y= -5 is a horizontal line. Since the problem says "x- 2, for x> 0", x=0 does NOT belong to this part of the grapy. Use an "open point" to show that. Since it says "-5, for x less than or equal to 0", x= 0 is included in this part of the graph. Use a "closed point" to show that.

Thank you