What is the vertices f(x)=2x^2-8x-5 ?

The vertices f(x)=2x^2-8x-5 is Minimum (2,-13)

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vertices f(x)=2x^2-8x-5

vertices

f (x) = 2 x^{2} - 8 x - 5

Minimum (2, - 13)

Solution steps

y = 2 x^{2} - 8 x - 5

The parabola parameters are:

a = 2, b = - 8, c = - 5

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} = - 13

Therefore the parabola vertex is

(2, - 13)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 2

Minimum (2, - 13)

f (x) = 2 x^{2} - 8 x + 6

y = + 0.5 x^{2} - 6 x + 21

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

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

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

What is the vertices f(x)=2x^2-8x-5 ?
The vertices f(x)=2x^2-8x-5 is Minimum (2,-13)