What is the vertices x^2-4x-5 ?

The vertices x^2-4x-5 is Minimum (2,-9)

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vertices x^2-4x-5

vertices

x^{2} - 4 x - 5

Minimum (2, - 9)

Solution steps

y = x^{2} - 4 x - 5

The parabola parameters are:

a = 1, b = - 4, c = - 5

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} = - 9

Therefore the parabola vertex is

(2, - 9)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 1

Minimum (2, - 9)

x^{2} - 7

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

f (x) = - 3 x^{2} + 4

y^{2} - 12 y + 4 x + 4 = 0

2 x - y^{2} = 4