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Help pls
part b for Q11
part a & b for Q12

Help pls part b for Q11 part a & b for Q12-example-1
User Tom Hanley
by
3.7k points

2 Answers

8 votes

Answer:

x=-1.56,2.56

User Aleung
by
4.2k points
12 votes

Answer:

x = 2.6, x = -1.6

Explanation:

If we compare the graph to points (-1, 2) and (2, 2) [see blue lines on attachment], we can determine that the scale of the y-axis is 1 square per 1 unit. (This is different from the scale of x-axis, which is 2 squares per unit).

Therefore, to use the graph to estimate the x-value when y = 4, count 4 square up the y-axis from 0, then trace along the horizontal line until you meet the curve. Then trace down to the x-axis and read the values of x. (shown by the red lines on the attachment).

So the approximate value of x is -1.6 and 2.6 when y = 4.

The actual answer can be calculated as follows:

x² - x = 4

⇒ x² - x - 4 = 0

Use the quadratic formula:


x=(-b\pm√(b^2-4ac))/(2a)


\implies x=(-(-1)\pm√((-1)^2-4(1)(-4)))/(2(1))


\implies x=(1\pm√(17))/(2)


\implies x=2.6,x= -1.6 \sf \ (1 \ dp)

Help pls part b for Q11 part a & b for Q12-example-1
User Tofig Hasanov
by
4.8k points