(a) Copy and complete the table of values for the relation \(y=2x^2-x-2\) for \(4 \le x \le 4\). x -4 -3 -2 -2 0 1 2 3 4 y 19 -2 26 (b) Using a scale of 2 c...
Question 1 Report
(a) Copy and complete the table of values for the relation \(y=2x^2-x-2\) for \(4 \le x \le 4\).
| x | -4 | -3 | -2 | -2 | 0 | 1 | 2 | 3 | 4 |
| y | 19 | -2 | 26 |
(b) 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 \(y=2x^2-x-2\) for \(4 \le x \le 4\).
(c) On the same axes, draw the graph of \(y=2x+3\).
(d) Use the graph to find the: (i) roots of the equation \(2x-3r-5\ 0\); (i) range of values of \(x\) for which \(2x^2-x-2<0\).
Answer Details
(a) Completed table for \(y = 2x^{2} - x - 2\), \(-4 \leq x \leq 4\). Substituting each value of \(x\):
\(x=-4:\ 2(16)+4-2=34\). \(x=-3:\ 2(9)+3-2=19\). \(x=-2:\ 2(4)+2-2=8\). \(x=-1:\ 2(1)+1-2=1\). \(x=0:\ -2\). \(x=1:\ 2-1-2=-1\). \(x=2:\ 8-2-2=4\). \(x=3:\ 18-3-2=13\). \(x=4:\ 32-4-2=26\).
| x | -4 | -3 | -2 | -1 | 0 | 1 | 2 | 3 | 4 |
|---|---|---|---|---|---|---|---|---|---|
| y | 34 | 19 | 8 | 1 | -2 | -1 | 4 | 13 | 26 |
(b) Graph. Plot the nine points (2 cm to 1 unit on the x-axis, 2 cm to 5 units on the y-axis) and join with a smooth upward (minimum) parabola; its lowest point is near \(x=0.25,\ y=-2.1\).
(c) Line \(y = 2x + 3\). Plot two points, e.g. \((0,3)\) and \((2,7)\), and draw the straight line on the same axes.
(d)(i) Roots of \(2x^{2} - 3x - 5 = 0\). Where the curve meets the line, \(2x^{2}-x-2 = 2x+3\), which rearranges to \(2x^{2}-3x-5 = 0\). So the x-coordinates of the two intersection points are the required roots. From the graph \(x = -1\) and \(x = 2.5\). (Check: \(2x^2-3x-5=(2x-5)(x+1)=0\Rightarrow x=2.5,\ -1\).)
(d)(ii) Range of \(x\) for which \(2x^{2} - x - 2 < 0\). This is where the curve lies below the x-axis (\(y<0\)). The curve crosses \(y=0\) at \(x=\frac{1\pm\sqrt{17}}{4}\), i.e. \(x \approx -0.8\) and \(x \approx 1.3\). Hence \(y<0\) between them:
\[-0.8 < x < 1.3.\]
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