The problem consists of three parts: (a) incomplete Table 2.2 for $y = 2x^2 - 4x - 3$, (b) drawing the graph of $y = 2x^2 - 4x - 3$ for $-2 \le x \le 6$ using a specified scale, and (c) finding the roots of the equation $2x^2 - 4x - 3 = 4$ and the gradient of the curve $y = 2x^2 - 4x - 3$ at $x = 4$ using the graph.

1. Problem Description

The problem consists of three parts: (a) incomplete Table 2.2 for $y = 2x^2 - 4x - 3$, (b) drawing the graph of $y = 2x^2 - 4x - 3$ for $-2 \le x \le 6$ using a specified scale, and (c) finding the roots of the equation $2x^2 - 4x - 3 = 4$ and the gradient of the curve $y = 2x^2 - 4x - 3$ at $x = 4$ using the graph.

2. Solution Steps

(a) Completing Table 2.2:

We need to find the values of $y$ for $x = 0, 1, 3, 4, 5$.

$y = 2x^2 - 4x - 3$

For $x = 0$:

$y = 2(0)^2 - 4(0) - 3 = -3$

For $x = 1$:

$y = 2(1)^2 - 4(1) - 3 = 2 - 4 - 3 = -5$

For $x = 3$:

$y = 2(3)^2 - 4(3) - 3 = 2(9) - 12 - 3 = 18 - 12 - 3 = 3$

For $x = 4$:

$y = 2(4)^2 - 4(4) - 3 = 2(16) - 16 - 3 = 32 - 16 - 3 = 13$

For $x = 5$:

$y = 2(5)^2 - 4(5) - 3 = 2(25) - 20 - 3 = 50 - 20 - 3 = 27$

Thus, the completed table is:

x | -2 | -1 | 0 | 1 | 2 | 3 | 4 | 5 | 6

---|----|----|----|----|----|----|----|----|----

y | 13 | 3 | -3 | -5 | -3 | 3 | 13 | 27 | 45

(b) Drawing the graph:

Scale: 2 cm to 1 unit on the x-axis and 2 cm to 5 units on the y-axis.

Plot the points from the completed table and draw a smooth curve.

(c) Using the graph:

(i) Roots of the equation $2x^2 - 4x - 3 = 4$:

Rewrite the equation as $2x^2 - 4x - 7 = 0$. We are given the graph of $y = 2x^2 - 4x - 3$. We want to solve $2x^2 - 4x - 3 = 4$, or $2x^2 - 4x - 7 = 0$. Let $y = 2x^2 - 4x - 3$. We are trying to solve $y = 4$. Therefore, the roots occur where the graph of $y = 2x^2 - 4x - 3$ intersects the line $y = 4$. From the graph, we find the $x$ coordinates of the intersection points. Using the graph, the approximate x values are -0.8 and 2.

8. (ii) Gradient of the curve $y = 2x^2 - 4x - 3$ at $x = 4$:

Draw a tangent to the curve at $x = 4$. Find two points on the tangent line, preferably far apart for accuracy. Let's use (3, 6) and (5, 26) as points from the tangent line near x=

4. Gradient = (change in y) / (change in x) = $(26 - 6) / (5 - 3) = 20 / 2 = 10$

3. Final Answer

(a) Completed Table 2.2:

x | -2 | -1 | 0 | 1 | 2 | 3 | 4 | 5 | 6

---|----|----|----|----|----|----|----|----|----

y | 13 | 3 | -3 | -5 | -3 | 3 | 13 | 27 | 45

(c)

(i) Roots of the equation $2x^2 - 4x - 3 = 4$: $x = -0.8, 2.8$ (approximately).

(ii) Gradient of the curve $y = 2x^2 - 4x - 3$ at $x = 4$: $10$.