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#1

$Old$ October 22nd, 2009, 03:22 PM

hawk123 $hawk123 is offline$

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$Default$ Question on Quadratic Application

A parabola whose equation is y=x2-2x+k has a turning point with coordinates (1,-5). Find the value of K

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#2

$Old$ October 22nd, 2009, 05:08 PM

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A parabola whose equation is y=x2-2x+k has a turning point with coordinates (1,-5). Find the value of K

$y(1) = -5$

$-5 = (1)^2 - 2(1) + k$

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#3

$Old$ October 22nd, 2009, 10:13 PM

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y
= x^2 - 2x + k
= x^2 - 2x + (-2/2)^2 - (-2/2)^2 + k
= x^2 - 2x + 1 - 1 + k
= (x^2 - 2x + 1) + (k - 1)
= (x - 1)^2 + (k - 1)
= (x - H)^2 + (K)
where (H,K) is the turning point = (1, -5)
hence -5 = k - 1
k = -4

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