Grade 9 Line Graphs Answers

These are answers on line graphs worksheets

1. 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)
2. b) c) d)
3. e) 3. a) m = 3 and the y-intercept = -6 b) m = -7 and the x-intercept = 3
4. c) m = -3 and the point (2; -6) d) m = 5 and the y-intercept = -8 e) m = 2 and the x-intercept = -5
5. 4. a) 𝑦 = 8𝑥 + 9 b) 𝑦 = −9𝑥 + 1 c) 𝑦 = −5𝑥 + 2
6. 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)
7. 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)
8. 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)
9. 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 1 −6−4 ∴ 𝑚 = 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)
10. 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)
11. 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)
12. 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