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NationalTutoringObservatory
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MathEd-PII

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student
so we write f(2) in the bix for X
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student
and then do -(2)+1
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volunteer
I don't know what"bix" means
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student
box
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student
so then we get 2,-1
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volunteer
Yes.
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volunteer
that gets a closed circle, right?
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volunteer
Now we move onto graphing the bottom equation.
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volunteer
The bottom equation starts at x =2, but it doesn't include 2, so that point will get an open circle.
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volunteer
If we use x =2 in the bottom equation, what do we get?
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student
ohh okay
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student
im a lil confused rn
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volunteer
what we are doing is graphing three lines
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volunteer
the top line is y =x+1 but only is valid for x <0
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volunteer
the middle line is y=-x+1 but we only graph it between x =0 and x =2
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student
ohh whats our other points
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student
that were using?
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student
are we still using 2 for x
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student
or what
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volunteer
The middle line stops at x =2 and the bottom lime starts at x=2, so you need to figure out f(2) for both of them
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volunteer
using the middle equation f(2) = -1
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volunteer
that gives us the point (2,-1) and gets a close circle
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student
wait Pat write it onmthe board
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[ { "pii_type": "PERSON", "surrogate": "Pat", "start": 5, "end": 8 } ]
volunteer
the bottom equation starts at x=2
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student
i think im confusinh myself
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volunteer
using x =2 in the bottom equation gives us 2- 1 =1
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student
ohh ok
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volunteer
so the bottom line starts at (2,1)
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student
anf its open
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student
ookkk
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volunteer
but that gets an open circle because it is x>2 for the bottom equation
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volunteer
what would f(4) equal?
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student
4-1=3 so 4,3
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volunteer
can you plot that on the graph?
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student
would it be open
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student
or closed
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volunteer
The question of whether a dot is open or closed is only for vaues of x where the graph shifts
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student
so closed
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volunteer
(4,3) is just a regular dot
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volunteer
all dots on a line are closed, except possibly at an endpoint.
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volunteer
Now you can just draw the line for x>2
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student
so will it be a straight like theu it
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volunteer
But your line cannot go lower than x =2. Your line starts at x =2
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volunteer
the lines cannot overlap
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student
would we add arrow or no!
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volunteer
yes, that's it, and you could add an arrow because it goes on to infinity in both directions
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student
like that?
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volunteer
yes. that's it/
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student
okk
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volunteer
Look at the graph. For values of x <0.we have graphed the line y =x+1
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volunteer
for valus of x between 0 and 2, we have graphed the line y=-x+1
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volunteer
and for values of x >2, we have graphed the line y=x-1
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volunteer
does that make sense now?
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student
wait can u write it on the board
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volunteer
write what on the board?
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student
wait
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student
so we idk
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student
im confused
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volunteer
The graph is on the board
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student
are we still on 9?
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volunteer
I believe we are finished with it. You have the graph for it.
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student
ohh okk
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student
so we on 10 now
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volunteer
for 10 they want you to tell them what mistake the person made
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student
yes
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volunteer
can you see the error? think about that.
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student
the second function is wrong
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volunteer
why do say that?
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volunteer
what is the second function?
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student
so the graph should have a closed circle at x=-1 and a open cricle for the first and second part
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student
the graph isnt showing
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student
like the correct
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student
points
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volunteer
You are right that the errors are with open and closed circles.
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volunteer
(-1,2) should be open
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volunteer
and (2,-2) should also be open
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volunteer
ok?
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volunteer
Hello?
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student
yesss
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student
mb i was writing it down
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volunteer
Do you think you get the basic idea of piecewise functions now?
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student
yes but i have 4 more to do
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student
i think im messing up w the grpahs
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volunteer
Sorry, but I need to go now. But you can post another request
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volunteer
have a good rest of the day.
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student
You too
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[ { "pii_type": "PERSON", "surrogate": "You", "start": 0, "end": 3 } ]
student
thank u sm for ur time
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volunteer
Hello
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student
Hello!
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student
Could I get some help with this question?
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volunteer
Jamie
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[ { "pii_type": "PERSON", "surrogate": "Jamie", "start": 0, "end": 5 } ]
volunteer
Do you have an idea of what you should do first?
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student
Multiply the 3 to the 6
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volunteer
That is sort of right.
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volunteer
what you are actually doing is multiply both sides by 3, right?
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student
Yes sorry
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volunteer
Whenever you have denominators in your problem, you should think of multiplying both sides to "clear" the denominators.
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volunteer
What's next here?
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student
That’s where I got stuck
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volunteer
what do they want you to solve for?
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