What is the vertices y=x^2-2x5 ?

The vertices y=x^2-2x5 is Minimum (5,-25)

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vertices y=x^2-2x5

vertices

y = x^{2} - 2 x 5

Minimum (5, - 25)

Solution steps

y = x^{2} - 2 x \cdot 5

Rewrite

y = x^{2} - 2 x 5

in the form

y = a x^{2} + b x + c

y = x^{2} - 10 x

The parabola parameters are:

a = 1, b = - 10, c = 0

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

x_{v} = - \frac{( - 10 )}{2 \cdot 1}

Simplify

x_{v} = 5

Plug in

x_{v} = 5

to find the

y_{v}

value

y_{v} = - 25

Therefore the parabola vertex is

(5, - 25)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 1

Minimum (5, - 25)

f (x) = - 4 (x - 2)^{2} + 4

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

- 2 x^{2} + 12 x - 14

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

- 2 x^{2} + 12 x