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

The vertices y=x^2-4x+7 is Minimum (2,3)

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vertices y=x^2-4x+7

vertices

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

Minimum (2, 3)

Solution steps

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

The parabola parameters are:

a = 1, b = - 4, c = 7

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

x_{v} = - \frac{( - 4 )}{2 \cdot 1}

Simplify

x_{v} = 2

Plug in

x_{v} = 2

to find the

y_{v}

value

y_{v} = 3

Therefore the parabola vertex is

(2, 3)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 1

Minimum (2, 3)

x = (y - 3)^{2}

y^{2} = x + 1

f (x) = - x^{2} + 8 x - 12

x^{2} - 6 x + 8

y = - 4 x^{2} - 24 x - 40