2023 WAEC Mathematics Theory a. Copy and complete the tables of values of y = \(2x^2 - x -...

a. Copy and complete the tables of values of y = \(2x^2 - x - 4\) for -3 ≤ x ≤ 3

x	-3	-2	-1	0	1	2	3
y	17			-4

b. Using a scale of 2 cm to 1 unit on the x-axis and 2 cm to 2 unit on the y-axis, draw the graph of y = \(2x^2 - x - 4\) for -3 ≤ x ≤ 3.

ci. Use the graph to find: the roots of the equation \(2x^2 - x - 4\)

ii. Use the graph to find the: values of x for which y increases as x increases;

iii. Use the graph to find the: minimum point of y.

b. the graph above, solution to question 8b.

ci. The roots of the equation from the graph are -1.2 and 1.7.

ii. Let y = \(2x^2 - x - 4\)

\(\frac{dy}{dx}\) = 4x - 1

At critical point \(\frac{dy}{dx}\) = 0

so that 4x - 1 = 0;

x = \(\frac{1}{4}\)

Using test points:

1. we use a test point less than \(\frac{1}{4}\), e.g x = 0

\(\frac{dy}{dx}\)(0) = 4(0) - 1 = -1

at this point, y = \(2x^2 - x - 4\) is decreasing

2. we use a test point greater than \(\frac{1}{4}\) e.g x = 1

\(\frac{dy}{dx}\)(1) = 4(1) - 1 = 4 - 1 = 3.

at this point, y = \(2x^2 - x - 4\) is increasing.

so, the function y = \(2x^2 - x - 4\) increases as x increases when x is greater than \(\frac{1}{4}\)

Therefore, the value of x for which y increases are x is greater than \(\frac{1}{4}\)

iii. The minimum point of y = -4.1.

{{ settings.no_comment_msg ? settings.no_comment_msg : 'There are no comments' }}