Using a scale of 2cm to 1 unit on the x- axis and 1cm to 1 unit on the y- axis, draw on the same axes the graphs of \(y = 3 + 2x - x^{2}; y = 2x - 3\) for \(-3 \leq x \leq 4\). Using your graph:
(i) solve the equation \(6 - x^{2} = 0\);
(ii) find the maximum value of \(3 + 2x - x^{2}\);
(iii) find the range of x for which \(3 + 2x - x^{2} \leq 1\), expressing all your answers correct to one decimal place.
\(y = 3 + 2x - x^{2}\) and \(y = 2x - 3\)
Table of values for the equation for \(-3 \leq x \leq 4\)
| x | -3 | -2 | -1 | 0 | 1 | 2 | 3 | 4 |
| \(y = 3 + 2x - x^{2}\) | -12 | -5 | 0 | 3 | 4 | 3 | 0 | 5 |
| \(y = 2x - 3\) | -9 | -7 | -5 | -3 | -1 | 1 | 3 | 5 |
(i) \(6 - x^{2} = 0\)
\(6 - 3 + 2x - x^{2} = 2x - 3\)
\(3 + 2x - x^{2} = 2x - 3\)
\(\therefore y = 2x - 3\)
Read the point where the two equations intersect on the graph.
x = -2.6 and x = 2.5.
(ii) Maximum value of \(3 + 2x - x^{2}\) is at y = 4.
(iii) Range for which \(3 + 2x - x^{2} \leq 1\) is represented by the shaded portion in the graph.
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