What is the vertices y=x^2-4x-1 ?

The vertices y=x^2-4x-1 is Minimum (2,-5)

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vertices y=x^2-4x-1

vertices

y = x^{2} - 4 x - 1

Minimum (2, - 5)

Solution steps

y = x^{2} - 4 x - 1

The parabola parameters are:

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

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} = - 5

Therefore the parabola vertex is

(2, - 5)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 1

Minimum (2, - 5)

(x - 4)^{2} + 2

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

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

y = 2 x^{2} + 15 x

f (x) = - x^{2} + 10 x - 16