# At what points does an object following a path defined by y = -4x^2 + 6x + 2 start and stop given that is always above the ground. The object follows a path where the height is defined in terms of the x-value as y = -4x^2 + 6x + 2

As the object always stays above the ground the points it starts from and stops are give by the root of the equation -4x^2 + 6x +...

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The object follows a path where the height is defined in terms of the x-value as y = -4x^2 + 6x + 2

As the object always stays above the ground the points it starts from and stops are give by the root of the equation -4x^2 + 6x + 2 = 0 as at the solution of the equation y = 0.

-4x^2 + 6x + 2 = 0

=> 2x^2 - 3x - 1 = 0

x1 = `(3 + sqrt(9 + 8))/4`

=> `3/4 + sqrt 17/4`

x2 = `3/4 - sqrt 17/4`

The object starts from and stops at the points where the value of x is `3/4 + sqrt 17/4` and `3/4 - sqrt 17/4`

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