Algebra

I need to find the vertex of the graph:

f(x)=2/5x^2-4x+14

I know that in order to find what x equals I need to solve -b/2a
which for this equation would be -(-4)/2(2/5) which got me 4/ (4/5)

which makes no sense!
So please help me understand what I did wrong!

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  1. I used Symbolab to find how to do this, and it said that -(-4)/2(2/5)
    equals 5

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  2. You just have to watch the order of operation

    in -b/(2a) , your a = 2/5 and b = -4
    so you want
    -(-4) / (2(2/5))
    = 4 / (4/5)
    = 4(5/4) , remember, when dividing by a fraction we instead multiply by the reciprocal of that fraction
    = 5

    so the x of the vertex is 5, sub this back into the original
    y = (2/5)(25) - 4(5) + 14
    = 10 - 20 + 14 = 4

    your vertex is (5,4)

  3. did you notice my set of brackets in
    -(-4)/(2(2/5)) ?

    you had -(-4)/2(2/5) , which changes the order of operation

  4. In vertex form, f(x) = a(x-h)+k where the vertex is at (h,k)
    f(x)=2/5x^2-4x+14
    = 2/5 (x^2 - 10x) + 14
    = 2/5 (x^2 - 10x + 25) + 14 - 2/5 * 25
    = 2/5 (x-5)^2 + 4
    So the vertex is at (5,4) as above

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