# Extreme pointFind the extreme point of f(x)=3-3x^2

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You may also use the alternative method, since the given function represents a quadratic function, whose graph is a parabola and its extreme point is the vertex of parabola.

Using the following formulas, you may evaluate the coordinates of the vertex, such that:

`x = -b/(2a) ; y = -(b^2 - 4ac)/(4a)`

You need to identify the leading coefficient `a = -3` and `c = 3` . Since there is no term that contains x, hence,` b = 0` .

`x = 0`

`y = (-4*(-3)*(3))/(4*(-3)) => y = -36/(-12) => y = 3`

You may sketch the parabola that opens downward, to visualize its extreme point that is the vertex `(0,3).`

**Hence, evaluating the vertex of the given function yields **`(0,3).`

The extreme values of a function can be found when the critical values are known.

The critical values are the roots of the 1st derivative.

We'll differentiate with respect to x, to determine the first derivative:

f'(x) = -6x

We'll cancel f'(x):

-6x = 0

x = 0

The given function has an extreme value for the critical value x = 0

Since the coefficient of x^2 is negative, then the function reaches the maximum values for x = 0

f(0) = 3 - 3*0^2 = 3

The maximum value of the function is f(0) = 3.