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

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

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vertices y=x^2-4x-5

vertices

y = 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)

y^{2} - 2 y - 8 x - 31 = 0

x^{2} - x - 6

x = 4 y^{2}

2 x^{2} - 3

x^{2} = - 12 y