What is the vertices y=2x^2-1-(5x-6) ?

The vertices y=2x^2-1-(5x-6) is Minimum (5/4 , 15/8)

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vertices y=2x^2-1-(5x-6)

vertices

y = 2 x^{2} - 1 - (5 x - 6)

Minimum (\frac{5}{4}, \frac{15}{8})

Solution steps

y = 2 x^{2} - 1 - (5 x - 6)

Rewrite

y = 2 x^{2} - 1 - (5 x - 6)

in the form

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

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

The parabola parameters are:

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

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

x_{v} = - \frac{( - 5 )}{2 \cdot 2}

Simplify

x_{v} = \frac{5}{4}

Plug in

x_{v} = \frac{5}{4}

to find the

y_{v}

value

y_{v} = \frac{15}{8}

Therefore the parabola vertex is

(\frac{5}{4}, \frac{15}{8})

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 2

Minimum (\frac{5}{4}, \frac{15}{8})

y = x^{2} - x - 20

\frac{- 3 x ^{2} - 2 x + 6}{5}

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

f (x) = 4 x^{2}

2 x^{2} + 2 x + 3 y = 0

What is the vertices y=2x^2-1-(5x-6) ?
The vertices y=2x^2-1-(5x-6) is Minimum (5/4 , 15/8)