1. Multiple Representations
for a relationship
This means
that one
Relationship: representation may
As hours worked increases, be converted into the
Mathematical allowance increases other representation.
Models Equation
A = 10 + 2H Words
‚My name’s Bobby. I get a
(On calc: y = 10 + 2x ) $10 allowance for the
month, & I can make
another $10 for every 5
hours of extra work I do!‛
Graph Table
Hours Allow-
Worked ance
0 10
5 20
10 30
15 40
Answering Questions
… …
Example: How many
40 90 hours should Bobby work
in order to buy a $37 video
game?

2. Starting with:
Words Equation Table Graph a description of some
real-world
situation
Words
My name’s Bobby. I get a $10
Equation allowance for the month, & I
Mathematical can make another $10 for every
Models Variables: Allowance depends on Hours Worked 5 hours of extra work I do!
dependent
variable
A = 10 + 2 ● H independent
variable
dep var start rate ind var 1. Identify the variables (what’s
changing?) and their relationship
3. 2. Rate of change =
Dep. var. chg = $10 = $2/hr
Graph Table Ind. var. chg 5 hr
Hours Allow- 3. Starting Allowance = $10
Worked ance
0 10
● 4. To make a table, pick #’s
5 20 for H, plug in to find A:
10 30 Ex: A = 10 + 2H
● A = 10 + 2*5 (5 hrs)
15 40 A = 10 + 10
… … A = $20
40 90
6. After plotting at least
two points, draw a line 5. Each row in the table is an (x,y) point
through the whole graph on the graph:
Ex: 10 hrs, $30 is the point: (10, 30)

3. Graph Table Equation Words
Words
Mathematical ‚The rubber band’s length is 3
Equation inches with no weight hanging on
Models it, and gets 0.5 inches longer for
each ounce added onto it.‛
Variables: Length depends on Weight
dependent independent
Starting with: variable
L = 3 + 0.5 ● W variable 6. Plug the two variables,
an experiment that dep var start rate ind var starting length, and rate of
measured the length change into an equation.
of a rubberband
with different Starting
5. Rate of change (slope of the graph) =
weights hung on it. Graph Length
Table ∆y = Dep. var. chg = 7-6 = 1 in. = 0.5 in
∆x Ind. var. chg 8-6 2 oz. oz
y
1. Make a Ind. Var (x) Dep. Var (y)
scatterplot 4. Starting Length = 3 in.
Weight Length
of the (oz.) (in.) With no weight (0 oz.), the
experimental band is 3 inches long.
data 0 3
The ‚y-intercept‛ on the
6 6 graph
∆x ∆y
2. Draw a 8 7
‚best fit‛
X
line through
the points
3. Estimate at least 2 points on the ‚best
fit‛ line, including at x=0 (a weight of 0
oz.), and place them into the table.
*Estimate points at a crosshairs (+)

4. Answering Questions with a graph
Equation:
Q1: If Bobby worked Graph A = 10 + 2H
9 hours, what would Q2: How many hours
y
his allowance be? should Bobby work in order
d to buy a $37 video game?
e
a) You’re given 9 hours
(independent variable ‘x’) d) You’re given $37
So go to 9 on the Hours axis c (dependent variable ‘y’)
So go to 37 on the Allow. axis
b) Go up till you hit the line
e) Go right till you hit the line
b
c) Go left till you hit the axis
f f) Go down till you hit the axis
You found the answer!
That’s the answer!
“If Bobby works 9 hour, his X “Bobby needs to work 13.5 hours
allowance will be $28.”
to buy a $37 video game.”
a
Answer Q1 with graphing calculator:
Answer Q2 with graphing calc:
1. In ‘Y=‘, put: y = 10 + 2x
1 – 3: same as for Q1 at left
2. Setup the WINDOW
4. Press TRACE, move  and 
(Xmin= 0, Xmax= 16, Xscl= 1,
until Y is close to 37
Ymin= 0, Ymax= 40, Yscl= 2)
5. It’s between these two points
3. Display the GRAPH
(9,28) 6. Try X=13.5 -- type: 13.5, Enter
4. Press ‘9’ (X=9), and Enter
(13.5, 37) So at X = 13.5 (hours),
5. Y=28 is your answer:
Y = 37 (allowance of $37)
So at X = 9 (hours),
Y = 28 (allowance of $28)

5. Answering Questions with a table
(on graphing calculator)
Words
Equation:
Table ‚The rubber band’s length is 3
L = 3 + 0.5W inches with no weight hanging on
Weight Length
(oz.) (in.) it, and gets 0.5 inches longer for
each ounce added onto it.‛
Q1: If the band is 0 3
stretched to 5.4 in., how 2 4
much weight is on it?
(to nearest 0.1 oz.) 4 5 Q2: If we attached a 3.7 oz.
weight to the band, how
6 6 long would it get?
1. Enter equation: y = 3 + 0.5x in Y=
2. Setup the table: TBLSET
*Always start X at TblStart=0, 1. Enter equation: y = 3 + 0.5x in Y=
and increase it by ∆Tbl=1 2. Setup the table: TBLSET
3. Display the TABLE *Start X at TblStart=0, and increase
it by ∆Tbl=1
4. We’re looking for 5.4 in the Y-
column. Hmm, it’s in between 5 3. Display the TABLE
(X=4) and 5.5 (X=5). 4. We’re looking for 3.7 in the X-
Let’s start a new table: TBLSET column. Hmm, not there!
Hey, since we know the X
Start it at TblStart=4, & increase (weight), can’t we just start there?
it by only ∆Tbl=.1 (since need the
nearest .1 oz.) Set TblStart=3.7 in TBLSET
5. Display the TABLE again, and 5. Display the TABLE again
scroll down till find Y = 5.4. X = 3.7 will be at the top
6. The X value (4.8) is your answer! 6. The Y value (4.85) is your answer!
The full answer is: The full answer is:
“If the band is stretched to 5.4 in., “If you hang a 3.7 oz. weight on the
then there’s 4.8 oz. hanging on it” band, it will stretch to 4.85 in.”