What is the vertices y=2x^2-8x+13 ?

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

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vertices y=2x^2-8x+13

vertices

y = 2 x^{2} - 8 x + 13

Minimum (2, 5)

Solution steps

y = 2 x^{2} - 8 x + 13

The parabola parameters are:

a = 2, b = - 8, c = 13

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} = 5

Therefore the parabola vertex is

(2, 5)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 2

Minimum (2, 5)

f (x) = - 2 x^{2} - 12 x - 16

f (x) = (x + 4)^{2} + 1

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

y = - x^{2} + 8 x - 15

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