What is the vertices 2x^2-8x+6 ?

The vertices 2x^2-8x+6 is Minimum (2,-2)

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vertices 2x^2-8x+6

vertices

2 x^{2} - 8 x + 6

Minimum (2, - 2)

Solution steps

y = 2 x^{2} - 8 x + 6

The parabola parameters are:

a = 2, b = - 8, c = 6

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

x_{v} = - \frac{( - 8 )}{2 \cdot 2}

Simplify

x_{v} = 2

Plug in

x_{v} = 2

to find the

y_{v}

value

y_{v} = - 2

Therefore the parabola vertex is

(2, - 2)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 2

Minimum (2, - 2)

f (x) = x^{2} + 14 x - 72

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

f (x) = x^{2} + 7

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

y = x^{2} + 12 x + 36