(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 ...
Assessment:WAEC SSCE - General Mathematics - 2005Subject:General Mathematics
(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).
Parabola through the tabulated points, with the line y = 2.5 used in part (c)(ii); it cuts the x-axis at x = -0.7 and x = 2.7, and is symmetrical about x = 1.
(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
\[ x = 1. \]
(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).
Parabola through the tabulated points, with the line y = 2.5 used in part (c)(ii); it cuts the x-axis at x = -0.7 and x = 2.7, and is symmetrical about x = 1.
(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
\[ x = 1. \]