What is the vertices y=x^2-10x-4 ?

The vertices y=x^2-10x-4 is Minimum (5,-29)

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vertices y=x^2-10x-4

vertices

y = x^{2} - 10 x - 4

Minimum (5, - 29)

Solution steps

y = x^{2} - 10 x - 4

The parabola parameters are:

a = 1, b = - 10, c = - 4

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

x_{v} = - \frac{( - 10 )}{2 \cdot 1}

Simplify

x_{v} = 5

Plug in

x_{v} = 5

to find the

y_{v}

value

y_{v} = - 29

Therefore the parabola vertex is

(5, - 29)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 1

Minimum (5, - 29)

x^{2} + 6 y = 0

f (x) = 7 x^{2} - x + 1

f (x) = 3 x^{2} - 6 x + 2

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

- x^{2} - 6 x - 9