What is the vertices (8x-23)^2-10 ?

The vertices (8x-23)^2-10 is Minimum (23/8 ,-10)

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vertices (8x-23)^2-10

vertices

(8 x - 23)^{2} - 10

Minimum (\frac{23}{8}, - 10)

Solution steps

y = (8 x - 23)^{2} - 10

Rewrite

y = (8 x - 23)^{2} - 10

in the form

y = a x^{2} + b x + c

y = 64 x^{2} - 368 x + 519

The parabola parameters are:

a = 64, b = - 368, c = 519

x_{v} = - \frac{b}{2 a}

x_{v} = - \frac{( - 368 )}{2 \cdot 64}

Simplify

- \frac{- 368}{2 \cdot 64} : \frac{23}{8}

x_{v} = \frac{23}{8}

Plug in

x_{v} = \frac{23}{8}

to find the

y_{v}

value

y_{v} = - 10

Therefore the parabola vertex is

(\frac{23}{8}, - 10)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 64

Minimum (\frac{23}{8}, - 10)

y = (x + 1)^{2} + 1

y = (x + 1)^{2} + 2

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

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

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

What is the vertices (8x-23)^2-10 ?
The vertices (8x-23)^2-10 is Minimum (23/8 ,-10)