Download Warm-up: 1. On his fifth birthday, Paul was 42” tall On his seventh

Thursday September 10, 2015
Warm-up:
1. On his fifth birthday, Paul was 42” tall
On his seventh birthday, he was 48” tall
Calculate the average rate of change in his
height?
2. The value of my new car after 2 years was
$11,200. When the car is 6 years old, the value
dropped to $6100. Find the average rate of
change in the car’s value?
Warm-up:
1. On his fifth birthday, Paul was 42” tall
On his seventh birthday, he was 48” tall
Calculate the average rate of change in his
height?
Solution:
(5, 42) and (7,48)
So, m = (48-42)/(7-5)
= 6/2
= 3”/year
2. The value of my new car after 2 years was
$11,200. When the car is 6 years old, the value
dropped to $6100. Find the average rate of
change in the car’s value?
Solution:
(2, $11,200) and (6, $6,100)
So, m = (11,200 -6,100)/(2-6)
= 5,100/-4
= -$1,275/year
Thursday September 10, 2015
2.2 Finding Slope and Rate of Change
Objective: To find slopes of lines and rates of change
EQ: How do you determine whether 2 non-vertical lines are
parallel or perpendicular
Example: What is the slope of the line thru (-2,1) and (3,5)
Solution:
Example: Without graphing, determine whether the line
through the given points rises, falls, is horizontal, or is vertical
a. (-5,1), (3,1) b. (-6,0), (2,-4) c. (-1,3), (5,8), d. (4.6), (4,-1)
Solution:
Example: Classify if the lines are parallel, perpendicular or
neither: Line 1: (-2,2), (0,-1)
Solution:
and Line 2: (-4,-1), (2,-3)
Friday September 11, 2015
Warm-up:
Identify the slope and y-intercept of the
equation: 2x – 5y = 15
Warm-up:
Identify the slope and y-intercept of the
equation: 2x – 5y = 15
Solution:
Solve for y:
2x – 5y = 15
- 5y = -2x + 15
y = (2/5) x – 3
Therefore,
m = 2/5
b = -3
Friday September 11, 2015
2.4 Write Equations of Lines
Objective: To write an equation of a line in different forms
EQ: How many ways can you write an equation of a line?
If given
Then
Slope (m) and y-intercept (b)
Use Slope-Intercept form:
y = mx + b
Slope (m) and a point (x1,y1 )
Use Point –Slope form:
y - y1 = m (x - x1 )
2 points (x1,y1 ) and (x1,y1 )
1. Find m
2. Use Point-Slope form
Example1: When slope (m) and y-intercept (b) are given
Write the equation for the line in the graph
Solution: Use Slope-Intercept form:
According to the graph, b = -1
and m = (-3 - -1) / (3 – 0) = -2/3
Therefore, the equation is y = (-2/3) x - 1
Example2: When slope (m) and a point (x1,y1) are given
Write an equation of a line line that passes thru (5,4) with m=-3
Solution:
Let (x1,y1 ) = (5,4) and use point-slope form:
= m (x - x1)
y - y1
y–4
= -3 (x – 5)
y–4
= -3x + 15
y
= -3x + 19
Example3: When slope 2 points (x1,y1) and (x2,y2) are given
Write an equation of a line line that passes thru (5,-2) and (2,10)
Solution:
1. Find its slope:
Let (x1,y1) and (x2,y2) = (5,-2) and (2,10)
then
2.
Use Point-Slope form to write the equation
Example4: Write equations of parallel or perpendicular lines
Write an equation of a line line that passes thru (-2, 3) and:
a. parallel to y1 = -4x + 1
b. perpendicular to y1 = -4x + 1
Solution:
a. If y2 // y1
then m2 = m1 = -4.
Use point-slope form
b. If y2 y1
then m2 = 1/4.
Use point-slope form
a. If y2 and y1 are perpendicular then m2 = -1/m1 = 1/4.
Use point-slope form
Monday September 14, 2015
Warm-up:
1. Graph y = x and y = 2x
2. Graph y = x and y = x +3
3. Graph the following:
a. y = 2
b. x = -3
Monday September 14, 2015
2.4 Graphing Equations of Lines
Objective: To graph linear equations in slope-intercept form
and standard form
EQ: Is it possible to graph a linear function using standard form
or slope-intercept form?
Steps to graph an equation using slope-intercept form
1. Write the equation in slope-intercept form (y=mx+b) by
solving for y
2. Identify the y-intercept (b). Use b to plot the point (0, b).
This is where the line crosses the y-axis
3. Identify the slope (m). Use m to plot the 2nd point
4. Draw a line through these 2 points
Example1: Graph 2y - 4 = - 4x
Solution:
1. Write the equation in slope-intercept form:
2y - 4 = - 4x
2y
= -4x + 4
y
= -2x + 2
2. Identify b and draw the 1st point
 Since b = 4, the 1st point is (0, 2)
3. Identify m and draw the 2nd point
 Since m = -2, the 2nd point is (1, 0)
Monday September 14, 2015
Steps to graph an equation using standard form
1. Write the equation in standard form (Ax + By = C)
2. Identify the x-intercept by setting y = 0 and solve for x.
The point is (x, 0) and this is where the line crosses x-axis
3. Identify the y-intercept by setting x = 0 and solve for y.
The point is (0, y) and this is where the line crosses y-axis.
4. Draw a line through these 2 points
Example1: Graph 3y - 12 = -2x
Solution:
1. Write the equation in standard form:
2. Identify x-intercept
 Let y = 0,
2x + 3(0) = 12
 Solve for x x
=6
3y - 12 = -2x
3y
= -2x + 12
2x + 3y = 12
 1st point is (6, 0)
3. Identify y-intercept
 Let x = 0,
2(0) + 3y = 12
 Solve for y y
=4
 2nd point is (0, 4)
4.
Draw a line thru these 2 points
How to recognize horizontal and vertical lines
1.
The graph of y = c is a horizontal line that goes thru (0, c)
2.
The graph of x = c is a vertical line that goes thru (c, 0)
Tuesday September 15, 2015
Warm-up:
Which ordered pair is a solution of 2x + 5y > 9?
a. (-4,-1)
b. (-2,3)
c. (2, -4)
d. (6, -1)
Warm-up:
Which ordered pair is a solution of 2x + 5y > 9?
a. (-4,-1)
b. (-2,3)
c. (2, -4)
d. (6, -1)
Solution:
Tuesday September 15, 2015
2.8 Graphing Linear Inequalities in Two Variables
Objective: To graph a linear inequality
EQ: What does a dashed boundary line on the graph of an
inequality represent?
Steps to graph a linear inequality:
1.
Graph the boundary line of the inequality.
Use dashed line for < or > and solid line for ≤ or ≥
2. Test a point that is not on the graph line to determine
if that point is part of the solution or not:
 If it is, shade the half-plane containing the point
 If not, shade the other half
Tips: use (0,0) as test point unless it lies on the graph.
Example1: Graph y > -2x
Solution:
1. Graph y = -2x.
 Use dashed line since the sign is >
2. Pick a point to test, say (1,1).
Substitute x value and y value of (1,1) to the inequality
Is it true that 1 > -2(1) ? Yes, (1 > -2)
then (1,1) is part of the solution
shade the half-plane that contains (1,1)
Example1: Graph 5x - 2y ≤ -4
Solution:
1. Graph 5x - 2y = -4
 Use solid line since the sign is ≤
2. Pick a point to test, say (0,0).
 Is 5(0) – 2(0) ≤ -4? No (0 – 0 ≤ -4)
 shade the half-plane that doesn’t contain (0,0)
Wednesday September 16, 2015
2.6 Drawing Scatter Plots and Best-Fitting Lines
Objective: To write an equation of a line using scatter points
EQ: Can you make an equation out of a bunch of scatter data?
Definition:
1. Scatter Plot is a graph of a set of data pairs (x,y)
 Positive correlation: y increases when x increases

Negative correlation: y decreases when x increases
Example1: When slope (m) and y-intercept (b) are given
Write the equation for the line in the graph
Solution: Use Slope-Intercept form:
According to the graph, b = -1
and m = (-3 - -1) / (3 – 0) = -2/3

Neither: no correlation between x and y
Therefore, the equation is y = (-2/3) x - 1
2. Correlation Coefficients (r) : r is a number that
measures how well x and y are correlated.
Wednesday September 16, 2015
2.6 Drawing Scatter Plots and Best-Fitting Lines
Steps to obtain a Best-Fitting Line:
1. Draw a scatter plot using given data
2. Sketch a line that follow most fit the given data points
(same number of points above and below the line)
3. Choose 2 points on the line to write an equation of the line
(these 2 points don’t have to be the original data)
Example: Find the best-fitting line, using the below table
Solution:
1. Use the above steps to plot
the data points and sketch a line
2. Choose 2 points to write an equation:
(1, 300) and (7, 548)
Steps to obtain a Best-Fitting Line using calculator:
1. STAT then 1:EDIT to enter data
2. STAT then CALC then 4:LinReg(ax+b) to obtain the equation
(STAT, CALC, 4:LinReg(ax+b), then L1, L2, Y1 to get the graph)
1. 2nd , STATPLOT to choose plot setup
2. ZOOM, 9 to plot