What is the vertices f(x)=2x^2-12x+1 ?

The vertices f(x)=2x^2-12x+1 is Minimum (3,-17)

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vertices f(x)=2x^2-12x+1

vertices

f (x) = 2 x^{2} - 12 x + 1

Minimum (3, - 17)

Solution steps

y = 2 x^{2} - 12 x + 1

The parabola parameters are:

a = 2, b = - 12, c = 1

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} = - 17

Therefore the parabola vertex is

(3, - 17)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 2

Minimum (3, - 17)

f (x) = x^{2} + 9 x + 8

8 x^{2} + 12 x + 3

- 5 x^{2} + 118 x + 7

y^{2} = - 6 x

4 x - 2 x^{2}

What is the vertices f(x)=2x^2-12x+1 ?
The vertices f(x)=2x^2-12x+1 is Minimum (3,-17)