Transcript 2-1 Graph 2 Variable Eqs

2-1: Graphing 2-Variable Equations
Objectives:
1. To graph equations
on coordinate axes
with labels and
scales
2. To use a graphing
calculator to draw a
scatter plot and a
best-fitting line
Assignment:
• P. 20: 15-20
• P. 27: 1-18
• Algebra II Textbook:
P.117-120: 1-6, 10, 12,
14, 20, 26, 29- 31
Rate of Change
A rate of change is how much one quantity
changes (on average) relative to another.
Slope can be used to represent
an average rate of change.
For slope, we measure how 𝑦 changes
relative to 𝑥.
Slope Definition
The slope m of a
nonvertical line is
the ratio of vertical
change (the ryse) to
the horizontal
change (the run).
ryse
ryse
Intercepts
6
The 𝒙-intercept of a
graph is where it
intersects the 𝑥-axis.
4
2
𝑎, 0
0, 𝑏
The 𝒚-intercept of a
graph is where it
intersects the 𝑦-axis.
y-intercept
-5
x-intercept
5
-2
Click me!
Slope-Intercept
Slope-Intercept Form of a Line:
If the graph of a line has slope 𝑚
and a 𝑦-intercept of (0, 𝑏), then
the equation of the line can be
written in the form 𝑦 = 𝑚𝑥 + 𝑏
Slope
𝑦-intercept
Slope-Intercept
To graph an equation in slope-intercept form:
Plot
2.
0, 𝑏
Solve
1.
for 𝑦
Draw
4.
line
Use 𝑚
to plot
3.
more
points
Standard Form
Standard Form of a Line
The standard form of a linear
equation is 𝐴𝑥 + 𝐵𝑦 = 𝐶, where
𝐴 and 𝐵 are not both zero.
Generally taken to be integers
Standard Form
To graph an equation in standard form:
1. = 0
Let 𝑥
2. = 0
Let 𝑦
Solve for 𝑦
Solve for 𝑥
This is the 𝑦intercept
This is the 𝑥intercept
3. line
Draw
Point-Slope Form
Given the slope and
a point on a line,
Point-Slope Form of a Line
you could easily
find the equation
A line through (𝑥1, 𝑦1) with
using the slopeslope m can be written in
intercept form.
the form 𝑦– 𝑦1 = 𝑚(𝑥– 𝑥1).
Alternatively, you
could use the pointslope form of a line.
Exercise 1
Graph each of the following lines
2
𝑦 =− 𝑥−4
3
5𝑥 − 2𝑦 = 15
𝑦−3=4 𝑥+4
Correlation
Let’s say a set of data
consists of two quantities,
𝑥 and 𝑦. In statistics, a
correlation exists between
𝑥 and 𝑦 if there is a linear
relation between 𝑥 and 𝑦.
If 𝑦 increases as 𝑥
increases, there is a
positive correlation.
Correlation
Let’s say a set of data
consists of two quantities,
𝑥 and 𝑦. In statistics, a
correlation exists between
𝑥 and 𝑦 if there is a linear
relation between 𝑥 and 𝑦.
If 𝑦 decreases as 𝑥
increases, there is a
negative correlation.
Correlation
Let’s say a set of data
consists of two quantities,
𝑥 and 𝑦. In statistics, a
correlation exists between
𝑥 and 𝑦 if there is a linear
relation between 𝑥 and 𝑦.
If there’s no obvious pattern,
there is approximately no
correlation.
Exercise 1
Describe the correlation shown by each scatter plot.
Correlation Coefficient
A correlation coefficient for a set of data
measures the strength of the correlation.
-1 ≤ r ≤ 1
Perfect negative
correlation = -1
No correlation = 0
Perfect positive
correlation = 1
Correlation Coefficient
A correlation coefficient for a set of data
measures the strength of the correlation.
-1 ≤ r ≤ 1
Exercise 2
The table shows the number 𝑦 (in
thousands) of alternative-fueled vehicles in
use in the United States 𝑥 years after
1997. Graph this data as a scatter plot.
Determine if a correlation exists.
Graphing Calculator Instructions
Entering the data:
1. Press the STAT key and choose 1:Edit…
2. Under the list L1, enter all of the 𝑥-values,
hitting ENTER between values.
3. Press the right arrow key, and under the
list L2, enter all of the 𝑦-values, hitting
ENTER between values.
4. Take few seconds to check for typos.
Graphing Calculator Instructions
Graphing the data:
1. Press 2ND then the Y= key to access the
STAT PLOT menu.
2. Choose your favorite Plot#, press ENTER.
3. Turn Plot On. Choose the scatter plot
icon. Make sure the 𝑥’s come from L1
and the 𝑦’s come from L2. Choose your
favorite Mark.
Graphing Calculator Instructions
Graphing the data:
4. Press the ZOOM key.
5. Choose 9:ZoomStat.
–
This chooses a good viewing rectangle
(domain and range) based on the values
entered in L1 and L2
6. Enjoy
Line of Best-Fit
If a strong correlation
exists between x
and y, where | r | is
near 1, then the
data can be
reasonable
modeled by a trend
line.
Line of Best-Fit
This line of best fit
lies as close as
possible to all the
data points, with as
many above as
below.
Click Me!
Exercise 3
Find the best-fitting line from Exercise 2.
Graphing Calculator Instructions
Finding the trend line:
1. Press 2ND then the 0 (zero) key to
access the CATALOG menu.
2. Press the “D” key (𝑥-1).
3. Scroll down and press ENTER on
DiagnosticOn. Press ENTER again.
–
You only have to do this once. Doing this will
return the r- and r2-values for a trend line.
Graphing Calculator Instructions
Finding the trend line:
4. Press the STAT key. Use the right arrow
key to move over to the CALC menu.
5. Choose 4:LinReg(ax+b).
–
This is called a linear regression, and it will
return the trend line in slope-intercept form.
6. Now you’re back on the HOME screen.
Now we have to tell it where to find the
data.
Graphing Calculator Instructions
Finding the trend line:
7. Press 2ND “1” for the L1 key. These are
your 𝑥-values.
8. Press the , (comma) key.
9. Press 2ND “2” for the L2 key. These are
your 𝑦-values.
10. Press the , (comma) key.
Graphing Calculator Instructions
Finding the trend line:
11. Press VARS key to access the variables
menu. Use the right arrow key to scroll
over to the Y-VARS menu.
12. Choose 1:Function…
13. Choose 1:Y1.
14. Press ENTER.
Graphing Calculator Instructions
Finding the trend line:
15. The HOME screen now shows you the
values for a (slope), b (y-intercept), r2
(r2), and r (the correlation coefficient).
16. To view the trend line, press the GRAPH
key.
Graphing Calculator Instructions
To evaluate a trend line at a data point:
17. On the HOME screen, input Y1 (VARS >
Y-VARS > Functions… > Y1)
18. In parentheses, enter your chosen 𝑥value: Y1(100) for example. Press
ENTER.
Exercise 4
Use your line of best fit to predict the
number of alternative-fueled vehicles in
use in the United States 14 years after
1997.
Exercise 5
The table gives the systolic blood pressure 𝑦
of patients 𝑥 years old. Determine if a
correlation exists. If it is a strong
correlation, find the line of best fit. Predict
the systolic blood pressure of a 16-yearold.
Exercise 9
According to your model
Y(16)=96, but:
Girls (Age 16)
Boys (Age 16)
122-132
125-138
Our prediction was so far
off because you can’t
reliably extrapolate this
far away from the data
2-1: Graphing 2-Variable Equations
Objectives:
1. To graph equations
on coordinate axes
with labels and
scales
2. To use a graphing
calculator to draw a
scatter plot and a
best-fitting line
Assignment:
• P. 20: 15-20
• P. 27: 1-18
• Algebra II Textbook:
P.117-120: 1-6, 10, 12,
14, 20, 26, 29- 31