(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...

Answer Details

(a) Completed table of values for \( y = 2x^2 - x - 2 \)

Substituting each value of \(x\) into \( y = 2x^2 - x - 2 \):

At \(x=-4:\; y = 2(-4)^2-(-4)-2 = 32+4-2 = 34\)
At \(x=-2:\; y = 2(4)-(-2)-2 = 8+2-2 = 8\)
At \(x=-1:\; y = 2(1)-(-1)-2 = 2+1-2 = 1\)
At \(x=1:\; y = 2(1)-1-2 = 2-1-2 = -1\)
At \(x=2:\; y = 2(4)-2-2 = 8-2-2 = 4\)
At \(x=3:\; y = 2(9)-3-2 = 18-3-2 = 13\)

(b) and (c) The graphs of \( y = 2x^2 - x - 2 \) and \( y = 2x + 3 \)

Using a scale of 2 cm to 1 unit on the \(x\)-axis and 2 cm to 5 units on the \(y\)-axis, the points from the table are plotted and joined with a smooth curve. The straight line \( y = 2x + 3 \) is drawn on the same axes (it passes through \((-4,-5)\) and \((4,11)\), with \(y\)-intercept \(3\)).

(d) Using the graph

(i) Roots of \( 2x^2 - 3x - 5 = 0 \).

Subtracting this equation from the plotted curve:

So the solutions of \( 2x^2 - 3x - 5 = 0 \) are given by the values of \(x\) where the curve \( y = 2x^2 - x - 2 \) meets the line \( y = 2x + 3 \). Reading off the two points of intersection on the graph:

(each to within \(\pm 0.1\)).

(ii) Range of values of \(x\) for which \( 2x^2 - x - 2 < 0 \).

This is the range where the curve lies below the \(x\)-axis, i.e. between the two points where the curve cuts the \(x\)-axis. Reading these cutting points from the graph:

(each end to within \(\pm 0.1\)).