SOLUTION: A right triangle has one vertex on the graph of y = 9 - x^2, x > 0 at (x,y) another at the origin, and the third on the positive x-axis at (x,0). Express the area A of the triangle

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Question 1208127: A right triangle has one vertex on the graph of y = 9 - x^2, x > 0 at (x,y) another at the origin, and the third on the positive x-axis at (x,0). Express the area A of the triangle as a function of x.
Answer by ikleyn(52855)   (Show Source): You can put this solution on YOUR website!
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A right triangle has one vertex on the graph of y = 9 - x^2, x > 0 at (x,y) another at the origin,
and the third on the positive x-axis at (x,0). Express the area A of the triangle as a function of x.
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Three given points are (0,0) (the origin), (x,0) and (x,9-x^2) with x > 0.


They form a right-angled triangle with the right angle at x-axis.


Notice that 2 cases are possible.


    (1)  0 < x < 3.    Then the triangle is in quadrant I.

    (2)  3 < x < .  Then the triangle is in quadrant IV.


In any case, the legs of this triangle are x units and |9-x^2| units (absolute value).


    At x= 3 the triangle is degenerated; so it is a special case, 
    which does not fit to the problem's formulation.


So, the area of this triangle is  A(x) = , x > 0, x =/= 3.     ANSWER

Solved.



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