precalc

f(x)=4x^2-8x+9

find the vertex the maximum the minimum the range and the intervals where it increases and decreases

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  1. Try completing the square first.
    4x^2 -8x +9 = 4(x^2 -2x +1) +9 -4
    = 4(x-1)^2 + 5

    The vertex is where it has its lowest value, which is obviously at x=1, f(x) = 5

    The function increases for x<1 and x>1
    There is no maximum.

    The range is 5 to infinity

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    posted by drwls

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