(a) (i) Using a scale of 2 cm to 1 unit on both axes, on the same graph sheet, draw the graphs of \(y – \frac{3x}{4} = 3\) and \(y + 2x = 6\).
(ii) From your graph, find the coordinates of the point of intersection of the two graphs.
(iii) Show, on the graph sheet, the region satisfied by the inequality \(y – \frac{3}{4}x \geq 3\).
(b) Given that \(x^{2} + bx + 18\) is factorized as \((x + 2)(x + c)\). Find the values of c and b.
Explanation
(a)(i) 
R is the required region.
(iii) Coordinates of the point of intersection of the two graphs = (1.1, 3.8)
(b) Assuming \(y - \frac{3x}{4} = 3\)
If x = 0, \(y - \frac{3}{4}(0) = 3 \implies y = 3\)
If y = 0, \(0 - \frac{3}{4}x = 3 \implies -3x = 12 ; x = - 4\)
| x | 0 | -4 |
| y | 3 | 0 |
Assuming \(y + 2x = 6\)
If x = 0, \(y + 2(0) = 6 \implies y = 6\)
If y = 0, \(0 + 2x = 6 \implies 2x = 6 ; x = 3\)
| x | 0 | 3 |
| y | 6 | 0 |
(b) \((x + 2)(x + c) = x^{2} + 2x + cx + 2c\)
\(x^{2} + (2 + c)x + 2c = x^{2} + bx + 18\)
\( 2c = 18 ; c = 9\)
\(b = 2 + c = 2 + 9 = 11\)
(b, c) = (11, 9).