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Quadractics in factored form finding the vertex

Algebra
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y=(x-5)2+4

Apr 24th, 2015

y = (x - 5)^2 + 4

Now simplifying

y = x^2 - 10x + 25 + 4

y = x^2 - 10x + 29

Where    a = 1     b = - 10     c = 29

Now

Co-ordinates of vertex are (h, k), then

h = − b/2a

h = - (- 10) / 2(1)

h = 10 / 2

h = 5

k = c - b^2 / 4a

k = 29 - (-10)^2 / 4(1)

k = 29 - 100/4

k = 29 - 25

k = 4

Hence the co-ordinate of the vertex are (h, k) =(5, 4)

Apr 24th, 2015

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Apr 24th, 2015
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Apr 24th, 2015
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