Worksheet 18 Memorandum: Graphs
Grade 9 Maths
1.
a)
c)
e)
2.
a)
i)
linear
ii)
b)
i)
non-linear
increasing
ii)
decreasing, then increasing
iii)
continuous
iii)
discrete
i)
non-linear
i)
non-linear
ii)
decreasing
ii)
decreasing
iii)
continuous
iii)
discrete
i)
non-linear
i)
linear
ii)
increasing
ii)
decreasing
iii)
continuous
iii)
continuous
d)
f)
b)
c)
d)
e)
3.
a)
m = 3 and the y-intercept = -6
b)
m = -7 and the x-intercept = 3
c)
m = -3 and the point (2; -6)
d)
m = 5 and the y-intercept = -8
e)
m = 2 and the x-intercept = -5
4.
a)
𝑦 = 8𝑥 + 9
b)
𝑦 = −9𝑥 + 1
c)
𝑦 = −5𝑥 + 2
5.
d)
𝑦 =𝑥−6
a)
i)
The gradient
𝑦 −𝑦
𝑚 = 𝑥2−𝑥1
2
1
5−2
∴ 𝑚 = 1−5
∴𝑚= −
ii)
3
4
The equation of the straight line
𝑦 = 𝑚𝑥 + 𝑐
3
∴ 𝑦 = −4𝑥 + 𝑐
3
we have 𝑚 = − 4
Substitute in either point. We’ve used (1; 5)
3
∴ 5 = − 4 (1) + 𝑐
3
∴ 𝑐 = 54
3
3
This means that the equation of the straight line is 𝑦 = − 4 𝑥 + 5 4
iii)
The y-intercept
3
From iv) above the y-intercept is (0; 5 4)
iv)
The x-intercept
Make y = 0
3
3
∴ 0 = −4𝑥 + 54
3
3
3
∴ −5 4 = − 4 𝑥
(divide by − 4)
2
∴ 73 = 𝑥
2
The x-intercept is (7 3 ; 0 )
b)
i)
The gradient
𝑦 −𝑦
𝑚 = 𝑥2−𝑥1
2
1
5−3
∴ 𝑚 = 5−1
1
∴𝑚=2
ii)
The equation of the straight line
𝑦 = 𝑚𝑥 + 𝑐
1
∴ 𝑦 = 2𝑥 + 𝑐
1
and we have that 𝑚 = 2
Substitute in either point. We’ve used (5; 5)
1
∴ 5 = 2 (5) + 𝑐
1
∴ 𝑐 = 22
1
1
∴ The equation of the straight line is given by 𝑦 = 2 𝑥 + 2 2
iii)
The y-intercept
1
From ii) above, the y-intercept is (0; 2 2)
iv)
The x-intercept
Let y = 0
1
1
∴ 0 = 2𝑥 + 22
1
1
∴ −2 2 = 2 𝑥
∴ −5 = 𝑥
The x-intercept is (-5; 0)
c)
i)
The gradient
𝑦 −𝑦
𝑚 = 𝑥2−𝑥1
2
1
8−6
∴ 𝑚 = 8−(−6)
1
∴𝑚=7
ii)
The equation of the straight line
𝑦 = 𝑚𝑥 + 𝑐
1
∴ 𝑦 = 7𝑥 + 𝑐
1
and we have that 𝑚 = 7
Substitute in either point. We’ve used (-6; 6)
1
∴ 6 = 7 (−6) + 𝑐
6
∴ 𝑐 = 67
1
6
∴ The equation of the straight line is given by 𝑦 = 7 𝑥 + 6 7
iii)
The y-intercept
6
From ii) above, the y-intercept is (0; 6 7)
iv)
The x-intercept
Let y = 0
1
6
∴ 0 = 7𝑥 + 67
6
1
∴ −6 7 = 7 𝑥
∴ 𝑥 = −48
The x-intercept is (-48; 0)
d)
i)
The gradient
𝑦 −𝑦
𝑚 = 𝑥2−𝑥1
2
∴𝑚=
1
−6−4
7−3
1
∴ 𝑚 = −2 2
ii)
The equation of the straight line
𝑦 = 𝑚𝑥 + 𝑐
1
∴ 𝑦 = −2 2 𝑥 + 𝑐
1
and we have that 𝑚 = −2 2
Substitute in either point. We’ve used (-6; 7)
1
∴ 7 = −2 2 (−6) + 𝑐
∴ 𝑐 = −8
1
∴ The equation of the straight line is given by 𝑦 = −2 2 𝑥 − 8
iii)
The y-intercept
From ii) above, the y-intercept is (0; −8)
iv)
The x-intercept
Let y = 0
1
∴ 0 = −2 2 𝑥 − 8
1
∴ 8 = −2 2 𝑥
1
∴ 𝑥 = −3 5
1
The x-intercept is ∴ (−3 5 ; 0)
e)
i)
The gradient
𝑦 −𝑦
𝑚 = 𝑥2−𝑥1
2
∴𝑚=
1
17−(−9)
1−5
1
∴ 𝑚 = −6 2
ii)
The equation of the straight line
1
𝑦 = 𝑚𝑥 + 𝑐
and we have that 𝑚 = −6 2
1
∴ 𝑦 = −6 2 𝑥 + 𝑐
Substitute in either point. We’ve used (1; 17)
1
∴ 17 = −6 2 (1) + 𝑐
1
∴ 𝑐 = 23 2
1
1
∴ The equation of the straight line is given by 𝑦 = −6 2 𝑥 + 23 2
iii)
The y-intercept
1
From ii) above, the y-intercept is (0; 23 2)
iv)
The x-intercept
Let y = 0
1
1
∴ 0 = −6 2 𝑥 + 23 2
1
1
∴ −23 2 = −6 2 𝑥
8
∴ 𝑥 = 3 13
8
The x-intercept is (3 13 ; 0)
f)
i)
The gradient
𝑦 −𝑦
𝑚 = 𝑥2−𝑥1
2
1
16−(−9)
∴ 𝑚 = −1−(−10)
7
∴ 𝑚 = 29
ii)
The equation of the straight line
𝑦 = 𝑚𝑥 + 𝑐
7
∴ 𝑦 = 29𝑥 +𝑐
7
and we have that 𝑚 = 2 9
Substitute in either point. We’ve used (-1; 16)
7
∴ 16 = 2 9 (−1) + 𝑐
7
∴ 18 9 = 𝑐
7
7
9
9
∴ The equation of the straight line is given by 𝑦 = 2 𝑥 + 18
iii)
The y-intercept
7
From ii) above, the y-intercept is (0; 18 9)
iv)
The x-intercept
Let y = 0
7
7
0 = 2 9 𝑥 + 18 9
7
7
∴ −18 9 = 2 9 𝑥
19
∴ 𝑥 = −6 25