# how to figure out that vertex of parabola is extremeexplain

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You may also test the monotony of the function, hence, you need to consider the following example to a better understanding, such that:

`f(x) = x^2 - 3x + 2`

You need to evaluate the roots of quadratic equation, such that:

`x_(1,2) = (3+-sqrt(9 - 8))/2 => x_(1,2) = (3+-1)/2`

`x_1 = 2; x_2 = 1`

You need to evaluate the coordinates of vertex using the following formulas, such that:

`x_v = -b/(2a) => x_v = 3/2`

`y_v = (4ac - b^2)/(4a) => y_v = -1/4`

**You may notice that selecting any value for x, smaller or larger than `x = 3/2` , you will obtain a larger value than `y_v = -1/4` , hence, `f(x) > y_v` for **`x in (-oo,3/2)U(3/2,oo).`

The vertex of a parabola is an extreme since the x coordinate of the vertex represents the root of the 1st derivative of the parabola.

We'll take an example:

y=-2(x-2)^2-2

We'll differentiate both sides:

dy/dx = -4(x-2)

We'l put dy/dx = 0

-4(x-2) = 0

x - 2 = 0

x = 2

The vertex of the parabola is V(2;f(2)) = V(2 ; -2)