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preparing for the formative assessment
Good Day!
Schedule:
Revise chapter 3
Practise
1 / 32
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Slide 1:
Slide
Wiskunde
Middelbare school
havo, vwo
Leerjaar 2
This lesson contains
32 slides
, with
interactive quizzes
and
text slides
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Lesson duration is:
40 min
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Items in this lesson
Good Day!
Schedule:
Revise chapter 3
Practise
Slide 1 - Slide
What paragraph do you find the most difficult?
3.1 The graph of a linear function
3.2 Finding the equation of a line
3.3 linear relationships
3.4 The balance method
3.5 Solving equations
3.6 Applying equations
Slide 2 - Poll
3.1 The graph of a linear function
How to draw a graph of a linear function in the coordinate plane
Make a table with
at least 2
points.
Plot these points in the coordinate plane.
Draw a line through these points.
Don't forget to write the formula next to your graph!
Slide 3 - Slide
Example
Draw the graph of y = -3x + 2.
1.
x
0
1
y
2
-1
For x =
0
y = -3 ·
0
+ 2 = 0 + 2 = 2
y = 2
For x =
1
y = -3 ·
1
+ 2 = -3 + 2 = -1
y = -1
This gives us the points:
(0,2) and (1,-1)
Slide 4 - Slide
Example
2. Plot (0,2) and (1,-1)
Slide 5 - Slide
Example
2. Plot (0,2) and (1,-1)
3. Draw a line through (0,2) and
(1,-1)
Slide 6 - Slide
Example
2. Plot (0,2) and (1,-1)
3. Draw a line through (0,2) and
(1,-1)
4.
Write the formula next to the
graph!!
Slide 7 - Slide
3.1 The graph of a linear function
How to check if a point is on a line
To check if point (
x
,
y
) in on a line you fill in the
x-coordinate
in the formula of the line and see if the answer matches the
y-coordinate
.
If the answer and the y-coordinate are the same, the point is on the line.
Slide 8 - Slide
Example
Calculate wether points A(
9
,
15
) and B(
-2
,-8) are on line y = 2x - 3
Check point A:
Filling in x =
9
in y = 2x-3 gives:
y = 2 ·
9
- 3 = 18 - 3 =
15
Check point B:
Filling in x =
-2
in y = 2x-3 gives:
y = 2 ·
-2
- 3 = -4 - 3 = -7
Point A(9,15) is on the line
Point B(-2,-8) is not on the line
Slide 9 - Slide
3.2 Finding the equation for a line
The formula for a linear relationship between x and y in an equation in the form
y = ax+b
Going 1 to the right means going
a
up
The line has y-intercept (0,
b
)
Slide 10 - Slide
y =
2
x
-3
Going 1 to the right means going
2
up
This line has y-intercept (0,
-3
)
Slide 11 - Slide
how to find the equation of a line
1. Assume y = ax + b
2. Find the y-intercept. This is (0,b)
3. Pick two grid points on the line.
4. Write down the equation.
a
=
r
u
n
r
i
s
e
Slide 12 - Slide
a = .....
Slide 13 - Open question
What is the
y-intercept?
A
(0,1)
B
(1,0)
C
(0,-1)
D
(-1,0)
Slide 14 - Quiz
What is the
equation of this line?
Answers from the previous questions:
a = -2
y-intercept = (0,-1)
Slide 15 - Open question
Equations of parallel lines
Lines with the same
gradient
are
parallel
Parallel
lines have the same
gradient
Line
l
and
k
both have gradient -2
Line
l
and
k
are
parallel
Line
m
has gradient 3
Line
m
is not parallel to line
l
or
k
Slide 16 - Slide
3.3 linear relationships
How to find the equation of a line where the axis have different lables than x and y.
The
horizontal
axis is labelled
p
and the
vertical
axis is labelled
N
The equation is in the form:
N
= a
p
+ b
Slide 17 - Slide
Find the equation
1. The equation is in the form
N = ap + b
2. The line goes through (0, -80), so b = -80
3.
4. N =
a
p +
b
,
b
= -80 and
a
= 5 so
N = 5p - 80
a
=
r
u
n
r
i
s
e
=
2
0
1
0
0
=
5
Slide 18 - Slide
The sum graph
To draw
the sum graph
of two graphs you need two points of this graph.
You can find these points in two way:
1. By adding the y-values of the two graphs with the same x-value
2. By finding the points directly above the x-intercepts of the two graphs
You don't need to know this for the test!
Slide 19 - Slide
1. Adding y-values
Given are the lines f and g with equations f: y = -2x + 6 and
g: y = 3x - 2
f.
g.
f + g.
Two points on the sum graph are :
(0,4) and (1,5)
x
0
1
y
6
4
x
0
1
y
-2
1
x
0
1
y
6 + -2
4 + 1
You don't need to know this for the test!
Slide 20 - Slide
You see the sum graph of line f and g. In what order would you do these steps to find the sum graph?
You don't need to know this for the test!
Step 1: find the x-intercept
Step 2:
Find the y-values directly above the y-intercepts
Step 3: Draw a line through the new points
Slide 21 - Drag question
1. substracting y-values
Given are the lines f and g with equations f: y = -2x + 6 and
g: y = 3x - 2
f.
g.
f - g.
Two points on the sum graph are :
(0,8) and (1,3)
x
0
1
y
6
4
x
0
1
y
-2
1
x
0
1
y
6 - -2
4 - 1
The difference graph
You don't need to know this for the test!
Slide 22 - Slide
1.
finding the y-values with the x-intercepts
Step 1: Find the x-intercept with the same
y-value
as the intersection point
Step 2: Find the x-intercept of the line you are substracting
Step 3: Find the point with the same x-value as the point you found in step 2 on the other line (in this case line f)
Step 4: Draw a line through the two points from step 1 and step 3
In this case we are trying to find lin f
-g
So we are looking for the x-intercept of line
g
You don't need to know this for the test!
Slide 23 - Slide
When you have an equation and you want to find the solution for x:
You can
Add
or
Substract
the same number on both sides
You can
multiply
or
divide
by the same number on both sides
Example:
5x + 7 = 32
-7 -7
5x + 7
-7
= 32
-7
5x + 0 = 25
5
x = 25
:5 :5
5x:
5
= 25:
5
x = 5
3.4 The balance method
We always want the variables on the lefthand side,
That's why we try to remove the 7 from the LHS first
Slide 24 - Slide
Try it yourself!
1. -7x + 3 = 24 2. -5 + 6x = 37
Try to remove the 3 on the left hand side first.
We only want x on the left hand side!
Hint
Try to remove the -5 on the left hand side first.
We only want x on the left hand side!
Hint
Slide 25 - Slide
Solutions
1. -7x + 3 = 24 2. -5 + 6x = 37
- 3 - 3 +5 +5
-7x + 3 - 3 = 24 - 3 -5 + 5 + 6x = 37 + 5
-7x + 0 = 21 0 + 6x = 42
-7
x = 21
6
x = 42
:-7 :-7 :6 :6
-7x : -7 = 21:-7 6x : 6 = 42 : 6
x = -3 x = 7
Slide 26 - Slide
3.5 Solving equations
In what order should you solve an equation (with brackets)?
1.
2.
3.
4.
If there are brackets, multiply them out.
Simplify the LHS and the RHS if needed.
Move all the terms with a variabel to the LHS and move all the terms without a variabel to the RHS.
Divide by the number in front of the variabel
Slide 27 - Drag question
3.5 Solving equations
Equations with fractions
When you want to change a fraction into a whole number, you multiply the fraction by it's denominator because:
so what should
we do when
4
3
⋅
1
4
=
4
1
2
=
3
5
3
x
=
6
4/1 = 4
?
Slide 28 - Slide
How would you solve
?
5
3
x
=
6
Slide 29 - Open question
3.6 Applying equations
You can solve many problems using an equation!!
Problem:
In 4 years I will be 5 times as old as I was 16 years ago. How old am i now?
I want to know my age right now so we will call this x.
Right now I am x years old.
How old am I in 4 years?
How old was I 16 years ago?
What is equal to my age 16 years ago
times 5
?
Slide 30 - Slide
3.6 Applying equations
You can solve many problems using an equation!!
Problem:
In 4 years I will be 5 times as old as I was 16 years ago. How old am i now?
Right now my age is x
How old am I in 4 years?
x + 4
How old was I 16 years ago?
x - 16
What is equal to my age 16 years ago
times 5
?
(x - 16) · 5 = x + 4
Now solve for x to find my age!
Slide 31 - Slide
What can I work on right now?
Mixed exercises: p. 122-123
Diagnostic test: p.126-127
Revision: p. 128-129
Slide 32 - Slide
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