How do you graph this and know if it is continuous or not?

Continuous because there are no breaks.

The graph is included as an attachment.

Step-by-step explanation:

Alright we want to graph:

y=x-4 for x<2

y=-2x+2 for x=2 or x>2

So let's graph the first piece y=x-4 for x<2.

I'm going to plug in 2 for x:   y=2-4=-2.   So this one line is going to contain an open circle at (2,-2). I say open circle because we did not have that we could actually include x=2 here because it says for x less than 2.

Now we are going to enter in one more number less than 2....your choice.

Let's go with x=0. When you plug in 0 for x into y=x-4 you get y=0-4=-4. So our line is going to include (0,-4).

So we are going to graph the points (0,-4) and (2,-2) again where (2,-2) is an open circle.  Connect these points. You may extend your line left because we have x<2, but do not extend it right of x=2.

Let's look at the other piece now: y=-2x+2 for x=2 or x>2.

I'm going to plug in 2 for x:  y=-2(2)+2=-4+2=-2 so we are going to include the point (2,-2) on our line.  This was actually a point we used from above that we didn't want to include.  We do now want to include because of the x=2 in our inequality so the dot can now be filled in. We need one more point to graph this line.  Let's plug in a number greater than 2 since our inequality say x=2 or x>2.  You choose.  How about x=3?

y=-2(3)+2=-6+2=-3.   So we are going to include the point (3,-3). So starting at (2,-2) and going right to connect it to (3,-3), you could extend passed the (3,-3) to the right.

I will show you my graph.

There are no breaks in the "curve", so it is continuous for all real numbers.

step-by-step explanation:

jjj

i can't solve unless i know what the line is

step-by-step explanation:

welcome to/p>