(a) Copy and complete the table of values for the equation \(y = 2x^{2} - 7x - 9\) for \(-3 \leq x \leq 6\).
| x |
-3 |
-2 |
-1 |
0 |
1 |
2 |
3 |
4 |
5 |
6 |
| y |
|
13 |
|
-9 |
-14 |
|
-12 |
|
6 |
|
(b) Using scales of 2cm to 1 unit on the x- axis and 2cm to 4 units on the y- axis, draw the graphs of \(y = 2x^{2} - 7x - 9\) for \(-3 \leq x \leq 6\).
(c) Use the graph to estimate the :
(i) roots of the equation \(2x^{2} - 7x = 26\);
(ii) coordinates of the minimum point of y;
(iii) range of values for which \(2x^{2} - 7x < 9\).
(a) Completing the table for \(y = 2x^{2} - 7x - 9\). Evaluate at each x:
| x | -3 | -2 | -1 | 0 | 1 | 2 | 3 | 4 | 5 | 6 |
|---|
| y | 30 | 13 | 0 | -9 | -14 | -15 | -12 | -5 | 6 | 21 |
Sample working: at \(x=-1,\ y=2+7-9=0\); at \(x=2,\ y=8-14-9=-15\).
(b) Graph. Using 2 cm to 1 unit (x-axis) and 2 cm to 4 units (y-axis), plot and join with a smooth parabola whose lowest point is near \(x=1.75\).
(c)(i) Roots of \(2x^{2}-7x=26\). Since \(2x^{2}-7x = y+9\), this becomes \(y+9=26\Rightarrow y=17\). Draw \(y=17\) and read:
\[x \approx -2.26 \quad\text{and}\quad x \approx 5.76.\]
(c)(ii) Minimum point. The turning point is at \(x=\dfrac{7}{4}=1.75\), where \(y=2(1.75)^2-7(1.75)-9=-15.125\). Minimum point \(\approx (1.75,\ -15.1)\).
(c)(iii) Range for \(2x^{2}-7x<9\). This is \(y<0\). The curve is below the x-axis between its roots \(x=-1\) and \(x=4.5\) (since \(2x^2-7x-9=(2x-9)(x+1)\)). Hence
\[-1 < x < 4.5.\]
(a) Completing the table for \(y = 2x^{2} - 7x - 9\). Evaluate at each x:
| x | -3 | -2 | -1 | 0 | 1 | 2 | 3 | 4 | 5 | 6 |
|---|
| y | 30 | 13 | 0 | -9 | -14 | -15 | -12 | -5 | 6 | 21 |
Sample working: at \(x=-1,\ y=2+7-9=0\); at \(x=2,\ y=8-14-9=-15\).
(b) Graph. Using 2 cm to 1 unit (x-axis) and 2 cm to 4 units (y-axis), plot and join with a smooth parabola whose lowest point is near \(x=1.75\).
(c)(i) Roots of \(2x^{2}-7x=26\). Since \(2x^{2}-7x = y+9\), this becomes \(y+9=26\Rightarrow y=17\). Draw \(y=17\) and read:
\[x \approx -2.26 \quad\text{and}\quad x \approx 5.76.\]
(c)(ii) Minimum point. The turning point is at \(x=\dfrac{7}{4}=1.75\), where \(y=2(1.75)^2-7(1.75)-9=-15.125\). Minimum point \(\approx (1.75,\ -15.1)\).
(c)(iii) Range for \(2x^{2}-7x<9\). This is \(y<0\). The curve is below the x-axis between its roots \(x=-1\) and \(x=4.5\) (since \(2x^2-7x-9=(2x-9)(x+1)\)). Hence
\[-1 < x < 4.5.\]