What is the vertices y=2x^2-12x-2 ?

The vertices y=2x^2-12x-2 is Minimum (3,-20)

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vertices y=2x^2-12x-2

vertices

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

Minimum (3, - 20)

Solution steps

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

The parabola parameters are:

a = 2, b = - 12, c = - 2

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

x_{v} = - \frac{( - 12 )}{2 \cdot 2}

Simplify

x_{v} = 3

Plug in

x_{v} = 3

to find the

y_{v}

value

y_{v} = - 20

Therefore the parabola vertex is

(3, - 20)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 2

Minimum (3, - 20)

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

f (x) = 7 x^{3} + 3

\frac{x ^{4}}{x ^{2} + x - 6}

f (x) = 5 x

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

What is the vertices y=2x^2-12x-2 ?
The vertices y=2x^2-12x-2 is Minimum (3,-20)