What is the vertices y=2x^2-8x+4 ?

The vertices y=2x^2-8x+4 is Minimum (2,-4)

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vertices y=2x^2-8x+4

vertices

y = 2 x^{2} - 8 x + 4

Minimum (2, - 4)

Solution steps

y = 2 x^{2} - 8 x + 4

The parabola parameters are:

a = 2, b = - 8, c = 4

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} = - 4

Therefore the parabola vertex is

(2, - 4)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 2

Minimum (2, - 4)

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

4 (x + 2)^{2} + 5

x^{2} - 2 x - 80

y^{2} + 2 y - 4 x + 9 = 0

f (x) = 2 x^{2} + 5 x - 12