What is the vertices y=2x^2-12x+3 ?

The vertices y=2x^2-12x+3 is Minimum (3,-15)

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vertices y=2x^2-12x+3

vertices

y = 2 x^{2} - 12 x + 3

Minimum (3, - 15)

Solution steps

y = 2 x^{2} - 12 x + 3

The parabola parameters are:

a = 2, b = - 12, c = 3

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} = - 15

Therefore the parabola vertex is

(3, - 15)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 2

Minimum (3, - 15)

f (x) = x^{2} - 8 x + 15

f (x) = - x^{2} + 8 x

+ (y - 2) = 3 (x - 5)^{2}

y = (x + 1) (x + 2)

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

What is the vertices y=2x^2-12x+3 ?
The vertices y=2x^2-12x+3 is Minimum (3,-15)