you can draw a square inside another square by placing each vertex of the inner square on one side of the outer square. The large square in this diagram has side length 10 units

a) Determine the area of the inscribed square as a function of x ( X is from one corner of the large square to the vertex of the inner square - the shorter section of the side which has been divided by two by the inner square)
b)range for area function

* I know the answer for a is A(x) = 2x² - 20x + 100 ---Answer
and I know how to do it.
But how do you find b. Some one said that the value of X can only be from 0-5. Why is that?

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asked by Annoym
  1. b) let's look at
    A(x) = 2x^2 - 20x + 100
    This is an upwards parabola.
    I will assume you know how to find its vertex.
    In standard form you would get
    A(x) = 2(x-5)^2 + 50
    with vertex at (5,50)
    This means your smallest inner square has an area of 50 square units when x = 5, clearly at the midpoint of the original square.

    Look back at your diagram,
    it is obvious that the domain of our A(x) is
    0 < x < 10
    at x = 0, A(0) = 100
    (the inner square coincides with the outer square)
    at x = 5 , A(5) = 50 , our vertex (minimum)
    at x = 10, A(10) = 100, back to same inner as outer

    since range would be all the A(x)'s possible
    the range is

    50 < A(x) < 100

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    posted by Reiny

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