Question 1: Match up the
equation with its graph
Starter
y = x2
y = -x2 + 2
y = -x2 - 2
y = x2 - 2
y = x2 + 2
y = -x2
a)
e)
5
5
b)
-5
c)
Question 2: Write the gradients
of each of these lines
5
-5
5
-5
-5
5
10
d)
f)
-5
5
-5
-5
5
-10
Question 1: Match up the
equation with its graph
Starter
y = x2
y = -x2 + 2
y = -x2 - 2
y = x2 - 2
y = x2 + 2
y = -x2
a)
Question 2: Write the gradients
of each of these lines
5
1
4
b)
2
-5
5
-5
Gradient = -1
Gradient = 2
d)
5
4
e)
f)
5
-5
-5
c)
5
1
10
1
-5
5
2
-5
5
8
-5
Gradient = 1/4
-10
Gradient = -4
Gradient from an Equation
 Find the gradient of a straight line graph from its equation.
𝑦 = 1 − 2𝑥
Putting in form 𝒚 =
𝒎𝒙 + 𝒄:
𝒚 = −𝟐𝒙 + 𝟏
Gradient is -2
?
2𝑥 + 3𝑦 = 4
Putting in form 𝒚 =
𝒎𝒙 + 𝒄:
𝟑𝒚 = −𝟐𝒙 + 𝟒
𝟐 ? 𝟒
𝐲=− 𝒙+
𝟑
𝟑
Gradient is −
𝟐
𝟑
Gradient using two points
! Given two points on a line, the gradient is:
𝑐ℎ𝑎𝑛𝑔𝑒 𝑖𝑛 𝑦 (𝑅𝑖𝑠𝑒)
𝑚=
𝑐ℎ𝑎𝑛𝑔𝑒 𝑖𝑛 𝑥 (𝑅𝑢𝑛)
1, 4
5, 7
2, 2
(3, 10)
𝑚 = 3?
(8, 1)
?
𝑚 = −2
(−1, 10)
8
𝑚 = −?
3
Exercise 1
1
2
Determine the gradient of the lines
which go through the following
points.
a
3,5 , 5,11
b
−1,0 , 4,3
c
2,6 , 5, −3
d
4,7 , 8,10
e
f
g
h
1,1 , −2,4
3,3 , 4,3
4, −2 , 2, −4
−3,4 , 4,3
𝒎 = 𝟑?
𝟑
𝒎= ?
𝟓
𝒎 = −𝟑
?
𝟑
𝒎= ?
𝟒
𝒎 = −𝟏
?
𝒎 = 𝟎?
𝒎 = 𝟏?
𝟏
𝒎 = −?
𝟕
Determine the gradient of the lines
with the following equations:
a 𝑦 = 5𝑥 − 1
𝒎 = 𝟓?
𝒎 = −𝟏
b 𝑥+𝑦=2
?
𝒎=𝟐
c 𝑦 − 2𝑥 = 3
?
d 𝑥 − 3𝑦 = 5
e 2𝑥 + 4𝑦 = 5
f
2𝑦 − 𝑥 = 1
g 2𝑥 = 3𝑦 − 7
3*
𝟏
𝒎= ?
𝟑
𝟏
𝒎 = −?
𝟐
𝟏
𝒎= ?
𝟐
𝟐
𝒎= ?
𝟑
A line 𝑙1 goes through the points
(2,3) and 4,6 . Line 𝑙2 has the
equation 4𝑦 − 5𝑥 = 1. Which
has the greater gradient:
𝟑
𝟓
𝒎𝟏 =
𝒎𝟐 =
𝟐
? 𝟒
So 𝒍𝟏 has greater gradient.
Gradients on a curved graph
•
You have already seen how to calculate the gradient of a linear graph by dividing
the change in y by the change in x
•
Also known as ‘rise over run’
•
What if you were asked to calculate the gradient of this line?
 The problem on a curved line is that the gradient is
different depending on where on the line you are…
 On the left side, the gradient will be negative
 On the right side, the gradient will be positive
Line is downward
sloping so the
gradient is
negative
Line is upward
sloping so the
gradient is
positive
 However, we can use a technique to estimate the
gradient, using what you already know…
Gradients on a curved graph
•
Estimate the gradient of this line where x = 1
Draw a straight line that just
touches the curve and equal space
either side of where x = 1 is.
 This line is known as a tangent
to the curve
4
 You can calculate the gradient
of it like on a straight line
graph
 The value will be an estimate
of the gradient of the curve
at the given point
2
𝐺𝑟𝑎𝑑𝑖𝑒𝑛𝑡 =
𝑟𝑖𝑠𝑒
𝑟𝑢𝑛
𝐺𝑟𝑎𝑑𝑖𝑒𝑛𝑡 =
4
2
𝐺𝑟𝑎𝑑𝑖𝑒𝑛𝑡 = 2
Gradient of a Cubic Graph
f(x) = x3 + 3x2 + 2x
Find the gradient where
x = -1
𝐺𝑟𝑎𝑑𝑖𝑒𝑛𝑡 =
𝑟𝑖𝑠𝑒
𝑟𝑢𝑛
−4
𝐺𝑟𝑎𝑑𝑖𝑒𝑛𝑡 =
4
4
𝐺𝑟𝑎𝑑𝑖𝑒𝑛𝑡 = −1
4
Area under a Graph
Speed-Time graphs
How would you
find the distance
travelled in this
case?
Speed
(m/s)
width
length
Time (s)
Key Points:
The object is moving with a constant speed.
The area under this graph would be a rectangle.
Area = length x width
Speed-Time graphs
[Task 1]
Speed (m/s)
One I prepared earlier!
[Task 2]
Mr Carne’s
Max Speed
was
1.8 m/s
Time (s)
Area = ½ x Base x Height
How far did Mr Carne travel in the first 6 seconds?
Area = ½ x 6 x 1.8
Shape under the graph for the first 6 seconds
is a triangle. The area of this equals the distance
travelled in the first 6 seconds.
Area = 5.4 m
Speed-Time graphs
Speed (m/s)
One I prepared earlier!
Time (s)
Area = length x width
How far did Mr Carne travel when at a constant speed?
Area = 1.8 x 6
Shape under the graph for a constant speed is a
rectangle. The area of this equals the distance
travelled while travelling at a constant speed.
Area = 10.8 m
Speed-Time graphs
Speed (m/s)
One I prepared earlier!
Time (s)
We have calculated this.
We have calculated this.
What is the total distance travelled by Mr Carne?
The total distance travelled by Hamilton is equal to
the whole area underneath the graph.
We must now calculate
this and total all three
areas together.
Speed-Time graphs
Speed (m/s)
Another one I prepared earlier!
Time (s)
Area = ½ x Base x Height
How far did Mr Carne travel between 12 and 15 seconds?
Shape under the graph for the this period of time
is a triangle. The area of this equals the distance
travelled between 12 and 15 seconds.
Area = ½ x 3 x 1.8
Area = 2.7 m
Speed-Time graphs
Area = 5.4 m
Area = 10.8 m
Area = 2.7 m
Speed (m/s)
Another one I prepared earlier!
Time (s)
Total distance travelled = Total area under the graph
Total distance travelled = 5.4 + 10.8 + 2.7
Total distance travelled = 18.9 m