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

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

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vertices y=2x^2-8x

vertices

y = 2 x^{2} - 8 x

Minimum (2, - 8)

Solution steps

y = 2 x^{2} - 8 x

The parabola parameters are:

a = 2, b = - 8, c = 0

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} = - 8

Therefore the parabola vertex is

(2, - 8)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 2

Minimum (2, - 8)

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

y = 2 x^{2} - 6 x

f (x) = - \frac{1}{3} x^{2} - \frac{7}{6} x + 1

4 x^{2} + 4 x - 8

f (x) = - 2 x^{2} - 16 x - 31