Question 1 Report
(a) Copy and complete the table.
\(y = x^{2} - 2x - 2\) for \(-4 \leq x \leq 4\)
| x | -4 | -3 | -2 | -1 | 0 | 1 | 2 | 3 | 4 |
| y | 22 | -2 | 1 | 6 |
(b) Using a scale of 2 cm to 1 unit on the x- axis and 2 cm to 5 units on the y- axis, draw the graph of \(y = x^{2} - 2x - 2\).
(c) Use your graph to find : (i) the roots of the equation \(x^{2} - 2x - 2 = 0\) ; (ii) the values of x for which \(x^{2} - 2x - 4\frac{1}{2} = 0\) ; (iii) the equation of the line of symmetry of the curve.
(a) Completing the table. For each value of \(x\) we evaluate \(y = x^{2} - 2x - 2\) by building it up in rows:
| \(x\) | -4 | -3 | -2 | -1 | 0 | 1 | 2 | 3 | 4 |
| \(x^{2}\) | 16 | 9 | 4 | 1 | 0 | 1 | 4 | 9 | 16 |
| \(-2x\) | 8 | 6 | 4 | 2 | 0 | -2 | -4 | -6 | -8 |
| \(-2\) | -2 | -2 | -2 | -2 | -2 | -2 | -2 | -2 | -2 |
| \(y\) | 22 | 13 | 6 | 1 | -2 | -3 | -2 | 1 | 6 |
The three missing entries are \(y=13\) (at \(x=-3\)), \(y=1\) (at \(x=-1\)) and \(y=-3\) (at \(x=1\)).
(b) Graph of \(y = x^{2} - 2x - 2\). Plotting the nine points \((x,y)\) and joining them with a smooth curve gives the parabola below. The dashed horizontal line \(y = 2.5\) is used in part (c)(ii) and the vertical dashed line \(x = 1\) is the line of symmetry from part (c)(iii).
(c) Using the graph.
(i) Roots of \(x^{2} - 2x - 2 = 0\). These are the values of \(x\) where the curve cuts the \(x\)-axis, i.e. where \(y = 0\). The curve crosses the \(x\)-axis at \[ x \approx -0.7 \quad \text{and} \quad x \approx 2.7. \]
(ii) Values of \(x\) for which \(x^{2} - 2x - 4\tfrac{1}{2} = 0\). Rearranging so that the left side becomes our plotted expression: \[ x^{2} - 2x - 4\tfrac{1}{2} = 0 \;\Rightarrow\; x^{2} - 2x - 2 = 2\tfrac{1}{2} = 2.5. \] So we draw the horizontal line \(y = 2.5\) and read off where it meets the curve: \[ x \approx -1.3 \quad \text{and} \quad x \approx 3.3. \]
(iii) Equation of the line of symmetry. The lowest point (vertex) of the curve occurs at \(x = \dfrac{-(-2)}{2(1)} = 1\), where \(y = -3\). The curve is symmetrical about the vertical line through this point, so the line of symmetry is \[ x = 1. \]
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