How to find the turning points from these quadratic equations?

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1 Answer
Feb 12, 2018

d) -2(x-2)^2 + 2
(2,2)
{y: y<=2}
f) 6(x+0.5)^2 - 18.375
(-0.5, -18.375)
{y:y>=-18.375}

Explanation:

-2x^2+8x -6:

factorise:

-2(x^2-4x+3)

(-4)/2 = -2

x^2-4x + 4 = (x-2)^2

x^2 - 4x + 3 = (x-2)^2 - 1

-2(x^2-4x+3) = -2((x-2)^2-1)

=-2(x-2)^2 + 2

a(x-h)^2 + k = -2(x-2)^2 + 2

turning point: (-h,k), where x=h is the axis of symmetry.

(-h, k) = (2,2)

x= 2 is the axis of symmetry.

since the coefficient of x^2 is negative (-2), the graph opens to the bottom.

the point (-h, k) is therefore a maximum point.

since the maximum point is the highest possible, the range is equal to or below 2.

{y: y<=2}

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6x^2+3x-18:

factorised:

6(x^2+0.5x-3)

x^2+0.5x+0.0625 = (x+0.25)^2

x^2+0.5x-3 = (x+0.25)^2-3.0625

6((x+0.25^2)-3.0625) = 6(x+0.25^2) - 18.375

a(x-h)^2 + k = 6(x+0.25)^2 - 18.375

turning point: (-h,k), where x=-h is the axis of symmetry.

(-h, k) = (-0.25,-18.375)

x= -0.25 is the axis of symmetry.

since the coefficient of x^2 is positive (6), the graph opens to the top.

(-h, k), therefore, is a minimum point.

(-h, k) = (-0.25, -18.375)

since the minimum point is the lowest possible, the range is equal to or above -18.375.

{y:y>=-18.375}

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