What is the vertices y=x^2-x-20 ?

The vertices y=x^2-x-20 is Minimum (1/2 ,-81/4)

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vertices y=x^2-x-20

vertices

y = x^{2} - x - 20

Minimum (\frac{1}{2}, - \frac{81}{4})

Solution steps

y = x^{2} - x - 20

The parabola parameters are:

a = 1, b = - 1, c = - 20

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{81}{4}

Therefore the parabola vertex is

(\frac{1}{2}, - \frac{81}{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{81}{4})

\frac{- 3 x ^{2} - 2 x + 6}{5}

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

f (x) = 4 x^{2}

2 x^{2} + 2 x + 3 y = 0

y = 4 x - x^{2}

What is the vertices y=x^2-x-20 ?
The vertices y=x^2-x-20 is Minimum (1/2 ,-81/4)