Question 13 (12 marks):
(a) Complete Table 2 in the answer space, for the equation $y=\frac{30}{x}$  by writing down the values of y when x = 2 and x = 5.

(b) For this part of the question, use the graph paper. You may use a flexible curve ruler.
Using a scale of 2 cm to 1 unit on the x-axis and 2 cm to 5 units on the y-axis, draw the graph of $y=\frac{30}{x}\text{ for }0\le x\le 7.$  

(c) From the graph in 12(b), find
(i) the value of y when x = 2.6,
(ii) the value of x when y = 17.5.

(d) Draw a suitable straight line on the graph in 12(b) to find the values of x which satisfy the equation $\frac{30}{x}=-5x+30\text{ for }0\le x\le 7.$
State the values of x.

Answer:
Table 2


Solution:
(a)



(b)



(c) From the graph
(i) When x = 2.6; y = 11.5
(ii) When y = 17.5; x = 1.7

(d)
$\begin{array}{l}\text{Given,}\frac{30}{x}=-5x+30\\ y=-5x+30\\ \text{From the graph; }x=1.25\text{ and }4.75\end{array}$

Question 12 (12 marks):
(a) Complete Table 2 in the answer space, for the equation y = –x³ + 4x + 10 by writing down the values of y when x = –2 and x = 1.5.

(b) For this part of the question, use the graph paper. You may use a flexible curve ruler.
Using a scale of 2 cm to 1 unit on the x-axis and 2 cm to 10 units on the y-axis, draw the graph of y = –x³ + 4x + 10 for –3 ≤ x ≤ 4.

(c) From the graph in 12(b), find
(i) the value of y when x = –2.5,
(ii) the positive value of x when y = 4.

(d) Draw a suitable straight line on the graph in 12(b) to find the values of x which satisfy the equation x³ – 14x + 5 = 0 for –3 ≤ x ≤ 4.
State the values of x.

Answer:




Solution:
(a)
y = –x³ + 4x + 10
When x = –2
y = –(–2)³ + 4(–2) + 10
y = 8 – 8 + 10
y = 10

When x = 1.5
y = –(1.5)³ + 4(1.5) + 10
y = –3.375 + 6 + 10
y = 12.625

(b)




(c) From graph
(i) When x = –2.5; y = 16
(ii) When y = 4; x = 2.5

(d)
x³ – 14x + 5 = 0
–x³ + 14x – 5 = 0
–x³ + (4x + 10x) + (10 – 15) = 0
–x³ + 4x + 10 = –10x + 15
Thus, y = –10x + 15


From graph, the values of x are 0.35 and 3.5.

Question 11 (12 marks):
(a) Complete Table 12 in the answer space, for the equation y = –x² + 2x + 10 by writing down the values of y when x = –1 and x = 2.

(b) For this part of the question, use the graph paper. You may use a flexible curve ruler.
Using a scale of 2 cm to 1 unit on the x-axis and 2 cm to 2 units on the y-axis, draw the graph of y = –x² + 2x + 10 for –3.5 ≤ x ≤ 4.

(c) From the graph in 12(b), find
(i) the value of y when x = –1.5,
(ii) the positive value of x when y = 8.2.
[adinserter block="3"]

(d) Draw a suitable straight line on the graph in 12(b) to find the values of x which satisfy the equation 7 – x = x² for –3.5 ≤ x ≤ 4.
State the values of x.

Answer:



Solution:
(a)
y = –x² + 2x + 10
When x = –1
y = –(–1)² + 2(–1) + 10
y = –1 – 2 + 10
y = 7

when x = 2
y = –(2)² + 2(2) + 10
y = –4 + 4 + 10
y = 10

(b)



(c) From graph
(i) When x = –1.5; y = 4.6
(ii) When y = 8.2; x = 2.7

(d)
y = –x² + 2x + 10 ……. (1)
0 = –x² – x + 7 ………. (2)
(1) – (2): y = 3x + 3 



From graph, the values of x are –3.2 and 2.2.

Question 10:
(a) Complete the table in the answer space for the equation y = x³ – 13x + 18 by writing down the values of y when x = –2 and x = 3.

(b) For this part of the question, use graph paper. You may use a flexible curve rule.
By using a scale of 2 cm to 1 unit on the x-axis and 2 cm to 5 units on the y-axis, draw the graph of y = x³ – 13x + 18 for –4 ≤ x ≤ 4 and 0 ≤ y ≤ 40.

(c) From your graph, find
(i) the value of y when x = –1.5,
(ii) the value of x when y = 25.

(d) Draw a suitable straight line on your graph to find the values of x which satisfy the equation x³ – 11x – 2 = 0 for –4 ≤ x ≤ 4 and 0 ≤ y ≤ 40.

Answer:


Solution:
(a)
y = x³ – 13x + 18   

when x = –2,
y = (–2)³ – 13(–2) + 18
   = –8 + 26 + 18
   = 36

when x = 3,
y = (3)³ – 13(3) + 18
   = 27 – 39 + 18
   = 6


(b)



(c)
(i) From the graph, when x = –1.5, y = 34.
(ii) From the graph, when y = 25, x = –3.25, –0.55 and 3.85.


(d)
y = x³ – 13x + 18 ----- (1)
0 = x³ – 11x – 2 ----- (2)
(1) – (2) : y = –2x + 20

The suitable straight line is y = –2x + 20.
Determine the x-coordinates of the two points of intersection of the curve y = x³ – 13x + 18 and the straight line y = –2x + 20.


From the graph, x = –3.2, –0.2 and 3.4.

Question 9:
(a) Complete the table in the answer space for the equation y = 8 – 3x – 2x² by writing down the values of y when x = –4 and x = 2.

(b) For this part of the question, use graph paper. You may use a flexible curve rule.
By using a scale of 2 cm to 1 unit on the x-axis and 2 cm to 5 units on the y-axis, draw the graph of y = 5 – 8x – 2x² for –5 ≤ x ≤ 3 and –27 ≤ y ≤ 9.

(c) From your graph, find
(i) the value of y when x = –2.5,
(ii) the positive value of x when y = –16.

(d) Draw a suitable straight line on your graph to find the values of x which satisfy the equation 8 – 3x – 2x² = 0 for –5 ≤ x ≤ 3 and –27 ≤ y ≤ 9.

Answer:

x	–5	–4	–3.5	–2	–1	0	1	2	3
y	–27	r	–6	6	9	8	3	s	–19

Calculate the value of r and s.


Solution:
(a)
y = 8 – 3x – 2x²  
when x = –4,
r = 8 – 3(–4) – 2 (–4)²
= 8 + 12 – 32 = –12

when x = 2,
s = 8 – 3(2) – 2(2)²
= 8 – 6 – 8 = –6


(b)



(c)
(i) From the graph, when x = –2.5, y = 2.5
(ii) From the graph, when y = –16, positive value of x = 2.8


(d)
y = 8 – 3x – 2x² ----- (1)
0 = 5 – 8x – 2x²  ----- (2)
(1) – (2) : y = 3 + 5x → y = 5x +3
The suitable straight line is y = 5x +3.

Determine the x-coordinates of the two points of intersection of the curve y = 8 – 3x – 2x² and the straight line y = 5x +3.

x	–5	0
y = 5x + 3	22	3

From the graph, x = –4.5, 0.55.

Question 8:
(a) Complete the table in the answer space for the equation y = x³ – 4x – 10 by writing down the values of y when x = –1 and x = 3.

(b) For this part of the question, use graph paper. You may use a flexible curve rule.
By using a scale of 2 cm to 1 unit on the x-axis and 2 cm to 10 units on the y-axis, draw the graph of y = x³ – 4x – 10 for –3 ≤ x ≤ 4 and –25 ≤ y ≤ 38.

(c) From your graph, find
(i) the value of y when x = 2.2,
(ii) the value of x when y = 15.

(d) Draw a suitable straight line on your graph to find the values of x which satisfy the equation x³ – 12x – 5 = 0 for –3 ≤ x ≤ 4 and –25 ≤ y ≤ 38.

Answer:

x	–3	–2	–1	0	1	2	3	3.5	4
y	–25	–10		–10	–13	–10		18.9	38

Solution:
(a)
y = x³ – 4x – 10  
when x = –1,
y = (–1)³ – 4(–1) – 10
= –7

when x = 3,
y = (3)³ – 4(3) – 10
= 5

(b)




(c)
(i) From the graph, when x = 2.2, y = –8
(ii) From the graph, when y = 15, x = 3.4


(d)
y = x³ – 4x – 10 ----- (1)
0 = x³ – 12x – 5 ----- (2)
(1) – (2) : y = 8x – 5

The suitable straight line is y = 8x–5. Determine the x-coordinates of the two points of intersection of the curve y = x³ – 4x – 10 and the straight line y = 8x –5.

x	0	2
y = 8x – 5	–5	–11

From the graph, x = –0.45, 3.7.

Question 7:

x	–3	–2	–1	0	1	2	3	3.5	4
y	19	3	r	1	3	–1	s	–31.4	–51

Solution:

x	0	4
y = –10x + 10	10	–30

Question 3:
On the graph in the answer space, shade the region which satisfies all three inequalities y ≥ x – 2, y > –3x + 6 and y ≤ 4.

Answer:


Solution:

Question 4:
On the graph in the answer space, shade the region which satisfies all three inequalities y > x – 2, y ≤ –x + 6 and x ≥ 1.

Answer:


Solution:

Question 5 (3 marks):
On the graph in the answer space, shade the region which satisfies all three inequalities y ≥ –3x + 6, y > x + 1 and y ≤ 5.

Answer:


Solution:

Question 1:
On the graph in the answer space, shade the region which satisfies all three inequalities y > –2x + 8, y ≥ x + 1 and y ≤ 7.

Answer:



Solution:

Question 2:
On the graph in the answer space, shade the region which satisfies all three inequalities 2x + y ≥ 2, x + y < 6 and y ≥ 1.

Answer:



Solution: