GCSE maths question about the turning point of the graph.
please help!

The coordinates of the turning point are (2, -9)

Step-by-step explanation:

The coordinates of the turning points of the quadratic equation

y = ax² + bx + c, are (h, k), where

∵ The equation of the curve is y = x² + bx + c

→ By comparing it with the form above

∴ a = 1

∵ The point (0, -5) lies on the curve

→ Substitute x by 0 and y by -5 in the equation to find the value f c

∵ -5 = (0)² + b(0) + c

∴ -5 = 0 + 0 + c

∴ -5 = c

→ Substitute it in the equatin

∴ y = x² + bx - 5

∵ The point (5, 0) lies on the curve

→ Substitute x by 5 and y by 0 in the equation to find the value f c

∵ 0 = (5)² + b(5) - 5

∴ 0 = 25 + 5b - 5

→ Add the like terms in the right side

∴ 0 = 20 + 5b

→ Subtract 5b from bth sides

∵ 0 - 5b = 20 + 5b - 5b

∴ -5b = 20

→ Divide both sides by -5 to find b

∴ b = -4

→ Substitute it in the equatin

∴ y = x² - 4x - 5

∵ a = 1 and b = -4

→ Substitute them in the rule of h above t find it

∵ h = =

∴ h = 2

→ To find k, substitute x by 2 and y by k

∵ k = (2)² - 4(2) - 5

∴ k = 4 - 8 - 5

∴ k = -9

∴ The cordinates of the turning point are (2, -9)

hope it you

step-by-step explanation:

sin(7pi/6 - pi/3)  

= sin(7pi/6 - 2pi/6)  

= sin(5pi/6)  

= sin(pi - pi/6)  

= sin(pi/6)  

= 1/2  

sin 7pi/6 - sin pi/3  

= sin (pi + pi/6) - sin pi/3  

= -sin pi/6 - sin pi/3  

= -1/2 - sqrt(3)/2  

= -(1+sqrt(3))2

step-by-step explanation: