What is the vertices f(x)=x^2-x-2 ?

The vertices f(x)=x^2-x-2 is Minimum (1/2 ,-9/4)

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vertices f(x)=x^2-x-2

vertices

f (x) = x^{2} - x - 2

Minimum (\frac{1}{2}, - \frac{9}{4})

Solution steps

y = x^{2} - x - 2

The parabola parameters are:

a = 1, b = - 1, c = - 2

x_{v} = - \frac{b}{2 a}

x_{v} = - \frac{( - 1 )}{2 \cdot 1}

Simplify

x_{v} = \frac{1}{2}

Plug in

x_{v} = \frac{1}{2}

to find the

y_{v}

value

y_{v} = - \frac{9}{4}

Therefore the parabola vertex is

(\frac{1}{2}, - \frac{9}{4})

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 1

Minimum (\frac{1}{2}, - \frac{9}{4})

f (x) = 3 x^{2} - 2 x - 1

7 x^{2} - 40 x - 25

f (x) = (x - 3)^{2} - 1

f (x) = x^{2} + 6 x + 7

y = 2 x^{2} - 12 x + 23

What is the vertices f(x)=x^2-x-2 ?
The vertices f(x)=x^2-x-2 is Minimum (1/2 ,-9/4)