What is the vertices y=2x^2-19x+9 ?

The vertices y=2x^2-19x+9 is Minimum (19/4 ,-289/8)

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vertices y=2x^2-19x+9

vertices

y = 2 x^{2} - 19 x + 9

Minimum (\frac{19}{4}, - \frac{289}{8})

Solution steps

y = 2 x^{2} - 19 x + 9

The parabola parameters are:

a = 2, b = - 19, c = 9

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

x_{v} = - \frac{( - 19 )}{2 \cdot 2}

Simplify

x_{v} = \frac{19}{4}

Plug in

x_{v} = \frac{19}{4}

to find the

y_{v}

value

y_{v} = - \frac{289}{8}

Therefore the parabola vertex is

(\frac{19}{4}, - \frac{289}{8})

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 2

Minimum (\frac{19}{4}, - \frac{289}{8})

4 x^{2} - 20 x - 24 y + 97 = 0

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

y = (x - 5)^{2}

y = - x^{2} - 4 x - 2

x^{2} - 16

What is the vertices y=2x^2-19x+9 ?
The vertices y=2x^2-19x+9 is Minimum (19/4 ,-289/8)