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

The vertices (5x-8)^2 is Minimum (8/5 ,0)

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vertices (5x-8)^2

vertices

(5 x - 8)^{2}

Minimum (\frac{8}{5}, 0)

Solution steps

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

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

The parabola parameters are:

a = 25, m = \frac{8}{5}, n = \frac{8}{5}

x_{v} = \frac{m + n}{2}

x_{v} = \frac{( \frac{8}{5} ) + ( \frac{8}{5} )}{2}

Simplify

\frac{( \frac{8}{5} ) + ( \frac{8}{5} )}{2} : \frac{8}{5}

x_{v} = \frac{8}{5}

Plug in

x_{v} = \frac{8}{5}

to find the

y_{v}

value

y_{v} = 0

Therefore the parabola vertex is

(\frac{8}{5}, 0)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 25

Minimum (\frac{8}{5}, 0)

f (x) = - 3 x^{2} + 6 x - 2

- x^{2} + 7

- x^{2} - 4

f (x) = - (x - 3)^{2} + 2

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