SOLUTION: A right triangle has one vertex on the graph of y = 3 - x^2, x > 0, and (x, y), another at the origin, and the third on the positive x-axis at (x, 0). Express the area A of the tri

Question 9590: A right triangle has one vertex on the graph of y = 3 - x^2, x > 0, and (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 prince_abubu(198) (Show Source): You can put this solution on YOUR website!

The graph above is for . One of your vertex lies on the curve. I apologize for the fact that it's quite difficult to draw a triangle that meets what the problem is saying. However, I'll do my best to explain it in words.

You know that the base of the triangle will ALWAYS lie in the x-axis. Its length (the base) is ALWAYS the x, since you're measuring from the origin to wherever x is.

Now the height of your triangle will depend on where your x is, and what the function will do to it. Actually, the height of your triangle will be the . Now, hang on because this is where it'll get tricky. After , will have a negative value. The only thing that happens is that your triangle becomes upside-down, and it'll still have an area. We will force the to be positive by putting it inside absolute value: . This way, if , you'll have a positive height for your upside-down triangle.

So, the area of the triangle is . The base B is the x. The height is . Since the area depends on x and the function , the area of the triange becomes a function of x:

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