4MA1/2HR Jun 2020, Question 18: Plotting Curves from a Table of Values

Part (a) · 2 marks

Fill in the missing y values in the table for y = (1/2)x³ − 2x + 3 at the given x values.

1. 1Substitute x = −3 into y = (1/2)x³ − 2x + 3: (1/2)(−27) − 2(−3) + 3 = −13.5 + 6 + 3 = −4.5
2. 2Substitute x = −2: (1/2)(−8) − 2(−2) + 3 = −4 + 4 + 3 = 3
3. 3Substitute x = −1: (1/2)(−1) − 2(−1) + 3 = −0.5 + 2 + 3 = 4.5
4. 4Substitute x = 0: (1/2)(0) − 0 + 3 = 3
5. 5Substitute x = 1: (1/2)(1) − 2(1) + 3 = 0.5 − 2 + 3 = 1.5
6. 6Substitute x = 2: (1/2)(8) − 2(2) + 3 = 4 − 4 + 3 = 3
7. 7Substitute x = 3: (1/2)(27) − 2(3) + 3 = 13.5 − 6 + 3 = 10.5

Answer−4.5, 3, 4.5, 3, 1.5, 3, 10.5

Part (b) · 2 marks

Draw the graph of y = (1/2)x³ − 2x + 3 on the grid for x from −3 to 3.

1. 1Plot the points from part (a) on the grid: (−3, −4.5), (−2, 3), (−1, 4.5), (0, 3), (1, 1.5), (2, 3), (3, 10.5)
2. 2Draw a smooth curve through all the plotted points

AnswerSmooth curve passing through (−3, −4.5), (−2, 3), (−1, 4.5), (0, 3), (1, 1.5), (2, 3), (3, 10.5)

Part (c) · 2 marks

Draw a suitable straight line on the grid and use it to estimate the solution of (1/2)x³ − x + 4 = 0.

1. 1Rewrite equation: (1/2)x³ − x + 4 = 0
2. 2Rearrange to: (1/2)x³ − 2x + 3 = −x − 1
3. 3So draw the line y = −x − 1
4. 4Find two points on y = −x − 1: when x = 0, y = −1; when x = −3, y = 2
5. 5Plot and draw the line y = −x − 1 on the grid
6. 6Find the intersection point of the curve and the line
7. 7Read off the x-coordinate of the intersection
8. 8−2.3 to −2.4

Answer−2.3 to −2.4

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