document.write( "Question 169278: a right traingle has one vertex on thr graph y=x^2 at (x,y), ANOTHER AT THE origin, and the third on the (positive) y-axis at (0,y). Express the area A of the triangle as a function of x. \n" ); document.write( "

Algebra.Com's Answer #124910 by Mathtut(3670) $\"\"$ $\"About$

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since $\"y=x%5E2\"$ \r
\n" ); document.write( "\n" ); document.write( "A=1/2 b h--------> b=y and h is $\"sqrt%28x%5E2%29\"$ =x\r
\n" ); document.write( "\n" ); document.write( "so A=1/2 (y)(x)
\n" ); document.write( ":
\n" ); document.write( "as a function of x
\n" ); document.write( ":
\n" ); document.write( " $\"A=%281%2F2%29%28x%5E2%29%28x%29\"$
\n" ); document.write( " $\"A=%281%2F2%29%28x%5E3%29\"$ \n" ); document.write( "

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