What is the vertices (3x-4)(4x-3)-2x ?

The vertices (3x-4)(4x-3)-2x is Minimum (9/8 ,-51/16)

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vertices (3x-4)(4x-3)-2x

vertices

(3 x - 4) (4 x - 3) - 2 x

Minimum (\frac{9}{8}, - \frac{51}{16})

Solution steps

y = (3 x - 4) (4 x - 3) - 2 x

Rewrite

y = (3 x - 4) (4 x - 3) - 2 x

in the form

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

y = 12 x^{2} - 27 x + 12

The parabola parameters are:

a = 12, b = - 27, c = 12

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

x_{v} = - \frac{( - 27 )}{2 \cdot 12}

Simplify

- \frac{- 27}{2 \cdot 12} : \frac{9}{8}

x_{v} = \frac{9}{8}

Plug in

x_{v} = \frac{9}{8}

to find the

y_{v}

value

y_{v} = - \frac{51}{16}

Therefore the parabola vertex is

(\frac{9}{8}, - \frac{51}{16})

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 12

Minimum (\frac{9}{8}, - \frac{51}{16})

4 x^{2}

y = x^{2} - 2

x^{2} + 2 x - 3

4 x^{2} - 24 x + 34

f (x) = 6 x^{2} - 5

What is the vertices (3x-4)(4x-3)-2x ?
The vertices (3x-4)(4x-3)-2x is Minimum (9/8 ,-51/16)