Word Count : 326

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The given curve is (y^2(x+1)=x^2(3-x)). Since (y) appears only as (y^2), the curve is symmetric about the (x)-axis. Writing (y^2=\dfrac{x^2(3-x)}{x+1}), real points exist only when the right-hand side is non-negative. First consider intercepts. Putting (y=0) gives (x^2(3-x)=0), so (x=0) or (x=3). Thus the curve passes through ((0,0)) and ((3,0)). Putting (x=0) gives (y^2=0), so the only point on the (y)-axis is the origin. Next consider the domain of (x). The expression (\dfrac{x^2(3-x)}{x+1}\ge 0). Since (x^2\ge 0), the sign depends on ((3-x)/(x+1)). _________ ______ _____ _____ ___ ______ _____ ________ __________ ________.
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