What is the vertices 2x^2-4x+7 ?

The vertices 2x^2-4x+7 is Minimum (1,5)

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vertices 2x^2-4x+7

vertices

2 x^{2} - 4 x + 7

Minimum (1, 5)

Solution steps

y = 2 x^{2} - 4 x + 7

The parabola parameters are:

a = 2, b = - 4, c = 7

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

x_{v} = - \frac{( - 4 )}{2 \cdot 2}

Simplify

x_{v} = 1

Plug in

x_{v} = 1

to find the

y_{v}

value

y_{v} = 5

Therefore the parabola vertex is

(1, 5)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 2

Minimum (1, 5)

x^{2} - 5 x - 84

x^{2} + 4 x - 24 y - 92 = 0

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

f (x) = - x^{2} + 7 x + 18

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