Question

The equation of a curve is given as y=x^2+2-3x. The curve intersects the x axis at

Answer 1

y = x^2 + 2 - 3x

y = x^2 - x - 2x + 2


y = x(x-1) -2(x-1)

y = (x-1)(x-2)

The zeroes of y are 1 and 2.

So it will intersect x axis at (1,0) and (2,0)

Answer 2

Hey ..!!

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y = x^2 + 2 - 3x

y = x^2 - 3x + 2

y,= x^2 - 2x - 1x + 2

y = x ( x - 2 ) - 1 ( x - 2 )

y = x - 1 ( x - 2)

Hence it will intersects X - axis at ( 1 , 0 ) , ( 2 , 0 )

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