Download Report

Christmas Break Revision
MYP5
Christmas Break
Revision
ISÄ
Teacher : Vivi Apostolou
Christmas Break Revision
MYP5
Representing Relations and Functions
A relation is a mapping of input values with output values.
The set of input values is the domain and the set of output values is the range.
So in simple words:
A relation is a function provided there is exactly one output for each input..
How to write a function
OR we can write f(x)=2x-1
ISÄ
Teacher : Vivi Apostolou
Christmas Break Revision
MYP5
Relation examples
Domain and Range examples
In example no1
Domain : {-3,-2,-1,1,2}
Range : {-2,-1,0,1,2}
Question : Can you write the relation as ordered pairs? Is this relation a function?
In example no2
Domain : {x│-1≤x≤0} and the range is {y│0≤y≤1}
Or we could write Domain x [-1,0] and range y[0,1]
Or Domain : {x : -1 ≤ x ≤ 0} and the range is {y: 0 ≤ y ≤ 1}
ISÄ
Teacher : Vivi Apostolou
Christmas Break Revision
MYP5
In example no3
Domain : {x│xR} and the range is {y│y≤4}
Linear Functions
A function is a linear function if it is of the form y=mx+b (where m and b are constants).
We can also use function notation f(x)=mx+b
Slope (gradient) and rate of change
The slope (or gradient) of the nonvertical line passing through the points (x1,y1) and (x2,y2) is:
𝑦2 − 𝑦1 𝑟𝑖𝑠𝑒
=
𝑥2 − 𝑥1 𝑟𝑢𝑛
Greater absolute value of the slope indicates steeper liner
Classification of lines by slope (gradient)
If two lines are Parallel then they have the same slope (m1=m2)
If two lines are Perpendicular then their slopes are negative reciprocals of each other
m1m2= -1 or m1 = -(1/m2)
ISÄ
Teacher : Vivi Apostolou
Christmas Break Revision
MYP5
Forms of a linear equation
Slope intercept form: y = mx+b (slope : m ------ y intercept : b)
Standard form: Ax + By = C ( where A,B are not both 0)
Writing an equation of a line
1. If you are given the slope m and the y-intersept b use :
Slope intercept form: y = mx+b (slope : m ------ y intercept : b)
2. If you are given the slope and a point (x1,y1) use:
Point-Slope form: y-y1 = m(x-x1)
3. If you are given 2 points (x1,y1) and (x2,y2) :
Find the slope m and use the point slope form
Scatter plots and correlation
A scatter plot is a graph used to determine whether there is a relationship between paired
data.
If y tends to increase as x increases, then the paired data have a positive correlation
If y tends to decrease as x increases, then the paired data have a negative correlation.
Approximating a best fitting line
1. Draw a scatter plot of the data
2. Sketch the line that appears to follow most closely the pattern.
3. Choose 2 point on the line; estimate the coordinates of each point. These don’t have
to be original data points
4. Find an equation of the line that passes through the two points from Step 3. This
equation models the data.
ISÄ
Teacher : Vivi Apostolou
Christmas Break Revision
MYP5
Exercices
1. Which of the following represents a function?
ISÄ
Teacher : Vivi Apostolou
Christmas Break Revision
MYP5
2. For each of the following graphs find the domain and the range.
3. Find the slope of the line that passes through the points:
4. Find the gradient of each line segment:
5. Find the gradient of the line segment joining pairs of point:
a. (2,3) and (8,4)
b. (1, -3) and (4,7)
c. (3, -2) and (3, -6)
d. (-7, 4) and (2,0)
e. (-5,7) and (1,5)
(5,7) and (-1, 7)
g. (15, -1) and (-12, -3)
h. (0, -10) and (-2, -7)
f.
6. Find the equation of the straight line that has slope m = 4 and passes through the point
(–1, –6).
ISÄ
Teacher : Vivi Apostolou
Revision
MYP5
7. Find the equation of the line that passes through the points (–2, 4) and (1, 2).
8. Your bank account increases linearly each week. If after 20 weeks of work, your bank
account is at 560€, while after 21 weeks of work it is at 585€, find a way to express the
relationship between how much money you've earned and how many weeks you've
worked in slope-intercept form.
9. Find the equation of the line. Use either slope-intercept form or point-slope form.
Make sure that you show all of your work!
i. The line goes through the point (-2, -4) and is parallel to the equation y = -x + 5.
j. The line goes through the point (-1, 2) and is perpendicular to the equation y=
(1/4)x -5
k. The line that goes through the points (3, 1) and (-2, 6)
Piecewise Function Exercises
10. The garage rates you pay for the weekend are:
2€ per half an hour but
10€ maximum for 20 hours
a. Write and graph a piecewise function for the parking charges shown on the
sign?
b. What are the domain and range of the function?
11. You have a summer job that pays time and half for overtime. That is, if you work more
than 40 hours per week, your hourly wage for the extra hours is 1.5 times your normal
wage of 10€.
c. Write and graph a piecewise function that gives your weekly pay P in terms of
the number h of hours you work.
d. How much will you get paid if you work 45 hours?
 x 2  1, x  0
f  x  
f  3 and f  2 
4x  1 x  0 .
12. Find
for the function
ISÄ
Teacher : Vivi Apostolou
Revision
MYP5
13. A parent drives from home to the grocery store at 0.9 mile per minute for 4 minutes,
stops to buy snacks for the team for 2 minutes, and then drives to the soccer field at a
speed of 0.7 mile per minute for 3 minutes. Write a piecewise function for the
parent’s distance from home to the soccer field during this time.
14. Write an equation of the graph shown [answ :abs(x-3)+4]
15. Graph the equation y = - |𝑥 + 2| + 3
Linear Programming Exercises
16. The area of a parking lot is 600 square meters. A car requires 6 square meters. A bus
requires 30 square meters. The attendant can handle only 60 vehicles. If a car is
charged $2.50 and a bus $7.50, how many of each should be accepted to maximize
income? [Answ(50,10)]
17. The Cruiser Bicycle Company makes two styles of bicycles: the Traveler, which sells for
$300, and the Tourister, which sells $600. Each bicycle has the same frame and tires,
but the assembly and painting time required for the Traveler is only 1 hour, while it is 3
hours for the Tourister. There are 300 frames and 360 hours of labor available for
production. How many bicycles of each model should be produced to maximize
revenue? [Answ(270, 30)]
ISÄ
Teacher : Vivi Apostolou
Revision
MYP5
System of equations
Methods of solving systems of linear equations
1. Linear Equations: Solutions Using Substitution with Two Variables
To solve systems using substitution, follow this procedure:
1.
2.
3.
4.
Select one equation and solve it for one of its variables.
In the other equation, substitute for the variable just solved.
Solve the new equation.
Substitute the value found into any equation involving both variables and solve for the
other variable.
Example
Solve the following system of linear equations.
y = 5x -1
2y = 3x + 12
ISÄ
Teacher : Vivi Apostolou
Revision
ISÄ
MYP5
Teacher : Vivi Apostolou
Revision
MYP5
2. Linear Equations: Solutions Using Elimination with Two Variables
To solve systems using substitution, follow this procedure:
5. Arrange both equations in standard form, placing like variables and constants one above
the other.
6. Choose a variable to eliminate, and with a proper choice of multiplication, arrange so
that the coefficients of that variable are opposites of one another.
7. Add the equations, leaving one equation with one variable.
8. Solve for the remaining variable.
9. Substitute the value found in Step 4 into any equation involving both variables and
solve for the other variable.
Example
Solve the following system of linear equations.
2x + 2y = 4
6x + 3y = 15
ISÄ
Teacher : Vivi Apostolou
Revision
MYP5
3. Linear Equations: Solutions Using Determinants with Two Variables
To solve systems using determinants (Cramer’s rule), follow this procedure:
To solve this system, three determinants are created. One is called the denominator
determinant, labeled D; another is the x-numerator determinant , labeled Dx; and the third is the
y-numerator determinant , labeled Dy .
The answers for x and y are then :
𝑥=
𝐷𝑥
𝐷𝑦
,𝑦 =
𝐷
𝐷
Example
Solve the following system of linear equations.
2x + 2y = 4
6x + 3y = 15
The denominator determinant, D, is formed by taking the coefficients of x and y from the equations
written in standard form.
𝐷= |
2 2
| = 2 ∙ 3 − 2 ∙ 6 = 6 − 12 = −6
6 3
The x‐numerator determinant is formed by taking the constant terms from the system and placing them
in the x‐coefficient positions and retaining the y‐coefficients.
𝐷𝑥 = |
4 2
| = 4 ∙ 3 − 2 ∙ 15 = 12 − 30 = −18
15 3
The y‐numerator determinant is formed by taking the constant terms from the system and placing them
in the y‐coefficient positions and retaining the x‐coefficients.
2 4
𝐷𝑦 = |
| = 2 ∙ 15 − 4 ∙ 6 = 30 − 24 = 6
6 15
The answers for x and y are as follows:
𝑥=
𝑦=
𝐷𝑥
𝐷
=
−18
−6
=3
𝐷𝑦
6
=
= −1
𝐷
−6
Solution : (3, -1)
ISÄ
Teacher : Vivi Apostolou
Revision
MYP5
4. Linear Equations: Solutions Using Graphing with Two Variables
To solve using graphing, graph both equations on the same set of coordinate axes and see
where the graphs cross. The ordered pair at the point of intersection becomes the solution
Example
Solve the following system of linear equations.
4x + 2y = 6
2x - 5y = 16
We graph the equations
Here are two things to keep in mind:


ISÄ
Dependent system. If the two graphs coincide—that is, if they are actually two
versions of the same equation—then the system is called a dependent system, and its
solution can be expressed as either of the two original equations.
Inconsistent system. If the two graphs are parallel—that is, if there is no point of
intersection—then the system is called an inconsistent system, and its solution is
expressed as an empty set {}, or the null set, ⊘.
Teacher : Vivi Apostolou
Revision
MYP5
System of equations (Exercises)
1. Solve the system of equations
2. Solve the system of equations using the Cramer’s rule
3. Solve the system of equations
ISÄ
Teacher : Vivi Apostolou