The table is for the relation (y = px^{2} – 5x + q).
| x | -3 | -2 | -1 | 0 | 1 | 2 | 3 | 4 | 5 |
| y | 21 | 6 | -12 | 0 | 13 |
(a)(i) Use the table to find the values of p and q.
(ii) Copy and complete the table.
(b) Using scales of 2cm to 1 unit on the x- axis and 2 cm to 5 units on the y- axis, draw the graph of the relation for (-3 leq x leq 5).
(c) Use the graph to find :
(i) y when x = 1.8 ; (ii) x when y = -8.
Explanation
(a) (i) (y = px^{2} – 5x + q)
| x | -3 | -2 | -1 | 0 | 1 | 2 | 3 | 4 | 5 |
| y | 21 | 6 | -12 | 0 | 13 |
From the table,
y = -12 when x = 0, i.e. (-12 = p(0^{2}) – 5(0) + q)
(q = -12).
y = 0 when x = 4, i.e. (0 = p(4^{2}) – 5(4) + q)
(0 = 16p – 20 + q implies 20 = 16p + q …. (1) )
Substitute q = -12 in (1),
(20 = 16p – 12)
(20 + 12 = 16p implies p = frac{32}{16} = 2)
Equation : (y = 2x^{2} – 5x – 12 ).
(ii) Given the equation (y = 2x^{2} – 5x – 12). The table becomes
| x | -3 | -2 | -1 | 0 | 1 | 2 | 3 | 4 | 5 |
| y | 21 | 6 | -5 | -12 | -15 | -14 | -9 | 0 | 13 |
(b)
(c) From the graph,
(i) When x = 1.8, y = -14.5 ; (ii) When y = -8, x = -0.65 or 3.15 or 3.2.