Need help with this mathswatch question

The x-coordinate of the turning point is

Step-by-step explanation:

The form of the quadratic equation is y = ax² + bx + c, where

Let us use this rule to solve the question

∵ y = x² + bx + c

∵ a is the coefficient of x²

∴ a = 1

∵ The curve intersects the y-axis at point (0, -14)

∴ The y-intercept is -14

∵ c represents the y-intercept

∴ c = -14

→ Substitute the value of c in the equation

∴ y = x² + bx + -14

∴ y = x² + bx - 14

∵ The curve intersects the x-axis at point (2, 0)

→ Substitute x, y in the equation by 2, 0 to find b

∵ 0 = (2)² + b(2) - 14

∴ 0 = 4 + 2b - 14

→ Add the like terms

∴ 0 = 2b + (4 - 14)

∴ 0 = 2b - 10

→ Add 10 to both sides

∴ 0 + 10 = 2b - 10 + 10

∴ 10 = 2b

→ Divide both sides by 2 to find b

∴ 5 = b

∵ x-coordinate of the turning point is h

∵ h =

∴ h =

∴ The x-coordinate of the turning point is

what?

step-by-step explanation:

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