What is the vertices f(x)=2x^2-4x-6 ?

The vertices f(x)=2x^2-4x-6 is Minimum (1,-8)

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vertices f(x)=2x^2-4x-6

vertices

f (x) = 2 x^{2} - 4 x - 6

Minimum (1, - 8)

Solution steps

y = 2 x^{2} - 4 x - 6

The parabola parameters are:

a = 2, b = - 4, c = - 6

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

x_{v} = - \frac{( - 4 )}{2 \cdot 2}

Simplify

x_{v} = 1

Plug in

x_{v} = 1

to find the

y_{v}

value

y_{v} = - 8

Therefore the parabola vertex is

(1, - 8)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 2

Minimum (1, - 8)

f (x) = 2 x^{2} - 4 x - 3

2 x^{2} - 1 + 5 x

y = - x^{2} + 18 x - 82

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

f (x) = x^{2} - 4 x - 21

What is the vertices f(x)=2x^2-4x-6 ?
The vertices f(x)=2x^2-4x-6 is Minimum (1,-8)