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anonymous

  • one year ago

How do you graph y=x^2-4x+3 I need the vertex, axis of symmetry, y and x axis as well

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  1. phi
    • one year ago
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    one way to do this is "complete the square" so that you can write the equation in the form y=a(x-h)^2 +k see http://www.mathwarehouse.com/geometry/parabola/standard-and-vertex-form.php

  2. anonymous
    • one year ago
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    How do you find the vertex I'm still a little confused

  3. anonymous
    • one year ago
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    first complete the square x^2-4x+3 x^2-4x=-3 -4/2=-2 -2^2=4 so add four to both sides of the equation x^2-4x+4=1 now factor the left side of the equation to get (x-2)^2=1 subtract 1 from both sides (x-2)^2-1=0 This is now in the form (x-h)^2+k the vertex is (h,k) so for your equation the vertex is (2,-1) the axis of symmetry is x=k so x=-1

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