What is extrema point of function f(x)=x^2+x-6?

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The maximum or the minimum of a quadratic function occurs at vertex of parabola representing the quadratic.

You may evaluate the coordinate of the vertex, either using the formulas `(-b/(2a),(4ac-b^2/(4a)))` , or using the derivative of the function.

If you select to use the formulas, you need to identify the coefficients `a,b,c,` such that:

`a = 1, b = 1, c = -6`

You need to replace its values in formulas above, such that:

`x = -1/(2*1) => x = -1/2`

`y = (4*1*(-6) - 1^2)/(4*1) => y = (-24 - 1)/4 => y = -25/4`

If you select to use the derivative of the function, you need first to solve for x the equation `f'(x) = 0` , such that:

`f'(x) = 2x + 1 => 2x + 1 = 0 => x = -1/2`

Evaluating the y coordinate at `x = -1/2` , yields:

`f(-1/2) = (-1/2)^2 + (-1/2) - 6 => f(-1/2) = 1/4 - 1/2 - 6`

`f(-1/2) = (1 - 2 - 24)/4 => f(-1/2) = -25/4`

**Hence, evaluating the coordinates of the vertex of parabola, using one method, or the other, yields `x = -1/2, y = -25/4` .**

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