### 5 (a)
First, we complete the given table for the relation [tex]\( y = \frac{5}{2} + x - 4 \)[/tex]:
[tex]\[
\begin{array}{c|c|c|c|c|c|c|c|c}
x & -2.0 & -1.5 & -1.0 & -0.5 & 0 & 0.5 & 1 & 1.5 \\
\hline
y & -15.5 & -3.0 & -2.5 & 1 & 2.5 & -1.0 & -0.5 & 0.0 \\
\end{array}
\][/tex]
Note that some values of [tex]\( y \)[/tex] were already provided, and others were calculated.
### 5 (b)
To draw the graph of the relation [tex]\( y = \frac{5}{2} + x - 4 \)[/tex] for [tex]\( -2.0 \leq x \leq 2.0 \)[/tex], follow these steps:
1. Set up your coordinate system with the scales:
- On the X-axis, use a scale of 2 cm for each unit.
- On the Y-axis, use a scale of 2 cm for every 5 units.
2. Plot the points from the completed table:
| [tex]\(x\)[/tex] | [tex]\(y\)[/tex] |
|-------|---------|
| -2.0 | -15.5 |
| -1.5 | -3.0 |
| -1.0 | -2.5 |
| -0.5 | 1 |
| 0 | 2.5 |
| 0.5 | -1.0 |
| 1 | -0.5 |
| 1.5 | 0.0 |
3. Plot these points on the graph paper accordingly.
4. Draw a smooth line joining these points to represent the relation.
### 5 (c)
The maximum value of [tex]\( y \)[/tex] can be observed from the table, which is:
[tex]\[
\boxed{2.5}
\][/tex]
### 5 (d)
To find the roots of the equation [tex]\( 8x^2 - 2x - 5 = 0 \)[/tex]:
1. From the table, the maximum value is [tex]\( y = 2.5 \)[/tex]. However, for the roots of the given quadratic equation, we solve for [tex]\( x \)[/tex].
2. The roots of the equation [tex]\( 8x^2 - 2x - 5 = 0 \)[/tex] are:
[tex]\[
\boxed{\frac{1}{8} - \frac{\sqrt{41}}{8}, \quad \frac{1}{8} + \frac{\sqrt{41}}{8}}
\][/tex]
These steps provide a comprehensive solution and the necessary steps to answer each part of the problem effectively.
