Algebra

I have a graph here. The vertex is
(4,-4), then the points (3,-3) and
(5,-3), then the points on the x axis are (2,0) and (6,0), then the points (1,5) and (5,5). This is a parabola that opens up.

I'm asked to write a function for f(x) in standard form.

I'm a little stuck. Any help you can provide is appreciated. Thank you.

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asked by Student
  1. You should be familiar with the vertex form of a parabola as
    y = a(x-p)^2 + q , where (p,q) is the vertex,

    so we have:
    y = a(x-4)^2 - 4
    using any of the other given points will give us the a
    let's use the easiest, (2,0)

    0 = a(2-4)^2 - 4
    0 = 4a - 4
    a= 1

    y = (x-4)^2 - 4 is your equation.

    A quick mental check shows that all your other points satisfy this equation.

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    posted by Reiny
  2. (x-4)^2 = k (y+4) from vertex

    put in some other point like (2,0)
    (2-4)^2 = k (4)
    4 = 4 k
    k = 1

    so I propose (x-4)^2 = y + 4
    check with another given point like (5,5)
    Hey (5,5) should be (7,5) !!!Typo?
    Try (3,-3) and (5,-3)

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    posted by Damon
  3. Thank you!

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    posted by Student
  4. You are welcome :)

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    posted by Damon

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