2.5 SPM Practice (Long Questions)

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.