Higher January 2023 Paper 1R Q15
15
(a) Complete the table of values for \(\;y = x^3 - 3x + 2\)
(2)
| \(x\) | −2 | −1 | −0.5 | 0 | 1 | 1.5 | 2 |
|---|---|---|---|---|---|---|---|
| \(y\) | 4 | 3.4 | 0 | 0.9 |
(b) On the grid, draw the graph of \(\;y = x^3 - 3x + 2\;\) for values of \(x\) from −2 to 2
(2)
(c) By drawing a suitable straight line on the grid, use your graph to find an estimate for the solution of
\(2x^3 - 3x + 4 = 0\)
Give your answer correct to one decimal place. (3)
\(2x^3 - 3x + 4 = 0\)
Give your answer correct to one decimal place. (3)
| Scheme | Marks | ||||||||||||||||
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
| B2 | ||||||||||||||||
| (2) |
Notes
B2: (B1 for 1 or 2 correct)
| Scheme | Marks |
|---|---|
| correct curve | B2 |
| (2) |
Notes
B2: For correct smooth curve. (there is an overlay for the curve – check the line now for (c))
If not B2, then B1 for at least 5 points plotted correctly ft from table dep on B1 or B2 in (a)
| Scheme | Marks |
|---|---|
\(2x^3 - 6x + 4 = -3x\) or \(x^3 - 3x + 2 = -\dfrac{3}{2}x\) or \(y = -\dfrac{3}{2}x\) seen (allow \(-\dfrac{3}{2}x\)) | M1 |
\(y = -\dfrac{3}{2}x\) allow a correct line that intercepts with the curve eg of points on line (0, 0), (−1, 1.5), (−1.5, 2.25), (−2, 3) | M1 |
| Answer dependent on a correct line being drawn Answer: \((x =)\;-1.6\) | A1ft |
| (3) | |
| (7 marks) |
Notes
M1: a correct line that intercepts with the curve
(a correct line drawn implies M2)
A1ft: accept −1.6 or −1.7 or ft their curve/line intercept dep on a correct line being drawn
NB: if \(y\) value given as well then M2 only