Carcass wrote:
Which points lie on the graph of \(y=\frac{x^2}{x+1}\)?
❑ \((-3,-5)\)
❑ \((-2,-4)\)
❑ \((-1,-3)\)
❑ \((1, \frac{1}{2})\)
❑ \((3, 2*\frac{1}{2})\)
Here we need to put the values of x ,y from the given option to the given equ. \(y=\frac{x^2}{x+1}\)
only option B and D satisfy the equa.
Considering option B
x= -2 and y =-4
we have \(-4\) = \(\frac{(-2)^2}{(-2+1)}\)=\(-4\)
and in option D
x= 1 , y =1/2
therefore \(\frac{1}{2}\)= \(\frac{(1)^2}{(1+1)}\)=\(\frac{1}{2}\)
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