algebra

find the vertex and the axis of symmetry andf(x)=x^2-10x+21`
graph the function

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asked by mema
  1. there are many ways to solve for the vertex. i think the easiest is to first find the derivative and solve for x:
    f(x)=x^2-10x+21
    f'(x) = 0 = 2x - 10
    2x = 10
    x = 5
    now substitute this back to original:
    f(5) = 5^2 - 10*5 + 21
    f(5) = 25 - 50 + 21
    f(5) = -4
    thus vertex is at (5, -4).
    and axis of symmetry is x = 5.

    hope this helps~ :)

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    posted by Jai

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