# Determine the maximum of the function x^2-x-6=0. The vertex  is the maximum/minimum point of a parabola.

To find the vertex:

x^2-x-6=0

==> (x-1/2)^2-1/4-6=0

Notice if we eliminate the parentheses, the equation becomes:

x^2-x+1/4-1/4-6=0. Therefore, the value of the equation has not changed.

(x-1/2)^2-6.25=0.

The vertex of the equation is 1/2, -6.25

However, since the parabola is concave up,...

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The vertex  is the maximum/minimum point of a parabola.

To find the vertex:

x^2-x-6=0

==> (x-1/2)^2-1/4-6=0

Notice if we eliminate the parentheses, the equation becomes:

x^2-x+1/4-1/4-6=0. Therefore, the value of the equation has not changed.

(x-1/2)^2-6.25=0.

The vertex of the equation is 1/2, -6.25

However, since the parabola is concave up, the vertex gives only the minimum value. To find the maximum point of the function, we need certain range of x value. If there is no range, then the y-value of function goes to infinity, thus has no maximum point.

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