What is the vertices x^2-14x+49 ?

The vertices x^2-14x+49 is Minimum (7,0)

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vertices x^2-14x+49

vertices

x^{2} - 14 x + 49

Minimum (7, 0)

Solution steps

y = x^{2} - 14 x + 49

The parabola parameters are:

a = 1, b = - 14, c = 49

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

x_{v} = - \frac{( - 14 )}{2 \cdot 1}

Simplify

x_{v} = 7

Plug in

x_{v} = 7

to find the

y_{v}

value

y_{v} = 0

Therefore the parabola vertex is

(7, 0)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 1

Minimum (7, 0)

y = x^{2} - 12 x + 36

f (x) = 3 x^{2} + 1

y^{2} + 2 y + x + 5 = 0

x^{2} + 6 x + 8

y = 2 x^{2} + 8 x - 34