Question 1 Report
(a) Copy and complete the following table of values for \(y = 2x^{2} - 9x - 1\).
| x | -1 | 0 | 1 | 2 | 3 | 4 | 5 | 6 |
| y | -1 | -8 | -11 | 17 |
(b) Using a scale of 2cm to represent 1 unit on the x- axis and 2cm to represent 5 units on the y- axis, draw the graph of \(y = 2x^{2} - 9x - 1\).
(c) Use your graph to find the : (i) roots of the equation \(2x^{2} - 9x = 4\), correct to one decimal place ; (ii) gradient of the curve \(y = 2x^{2} - 9x - 1\) at x = 3.
(a) Completing the table for \(y = 2x^{2} - 9x - 1\).
| \(x\) | -1 | 0 | 1 | 2 | 3 | 4 | 5 | 6 |
|---|---|---|---|---|---|---|---|---|
| \(y\) | 10 | -1 | -8 | -11 | -10 | -5 | 4 | 17 |
(b) Plot the points and join with a smooth parabola.
(c)(i) Roots of \(2x^{2} - 9x = 4\). Since \(y = 2x^{2}-9x-1\), rewrite \(2x^{2}-9x-4=0\) as \(2x^{2}-9x-1 = 3\). Draw the line \(y = 3\) and read where it cuts the curve:
\[x = \frac{9 \pm \sqrt{81+32}}{4} = \frac{9 \pm 10.63}{4} \;\Rightarrow\; x \approx -0.4 \text{ or } 4.9\](c)(ii) Gradient at \(x = 3\). Draw a tangent at \(x=3\); its slope is
\[\frac{dy}{dx} = 4x - 9 = 4(3) - 9 = \mathbf{3}\]Everything you need to excel in your exams