2.5 Graph of Functions (II), SPM Paper 2 (Long Questions)


Question 7:
(a) The following table shows the corresponding values of x and y for the 
equation y= –x3 + 3+ 1. 
 
x
–3
–2
–1
0
1
2
3
3.5
4
y
19
3
r
1
3
–1
s
–31.4
–51
Calculate the value of r and s.
 
(b) For this part of the question, use graph paper. You may use a flexible curve rule.
By using a scale of 2cm to 1 unit on the x-axis and 2cm to 5 units on the y-axis, draw the graph of  y = –x3 + 3x + 1 for –3 ≤ x ≤ 4 and –51 ≤ y ≤ 19.

(c) 
From your graph, find
(iThe value of when x = –2.8,
(iiThe value of when y = 30.

(d) 
Draw a suitable straight line on your graph to find the values of x which satisfy the equation –x3 + 13x – 9 = 0 for –3 ≤ x ≤ 4 and –51 ≤ y ≤ 19.
 
Solution:
(a)
y= –x3 + 3+ 1 
when x = –1,
= – (–1)3 + 3(–1) + 1
  = 1 – 3 + 1 = –1
when x = 3,
= – (3)3 + 3(3) + 1 = –17
 
(b)


(c) 
(iFrom the graph, when x = –2.8, y = 15
(iiFrom the graph, when y = 30, x = 3.5

(d)
y= –x3 + 3x+ 1 —– (1)
x3+ 13x – 9 = 0 —– (2)
y= –x3 + 3+ 1 —– (1)
0 = –x3 + 13x – 9 —— (2) ← (Rearrange (2))
(1)  – (2) : y = –10x + 10
 
The suitable straight line is y = –10x + 10.
 
Determine the x-coordinates of the two points of intersection of the curve 
y = –x3 + 3+ 1 and the straight line y = –10x10.
 
x
0
4
y = 10x + 10
10
–30
From the graph, x= 0.7, 3.25.

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 2cm to 1 unit on the x-axis and 2cm 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 = 8x5. 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.

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