What is the vertices y=x^2-5x-6 ?

The vertices y=x^2-5x-6 is Minimum (5/2 ,-49/4)

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vertices y=x^2-5x-6

vertices

y = x^{2} - 5 x - 6

Minimum (\frac{5}{2}, - \frac{49}{4})

Solution steps

y = x^{2} - 5 x - 6

The parabola parameters are:

a = 1, b = - 5, c = - 6

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

x_{v} = - \frac{( - 5 )}{2 \cdot 1}

Simplify

x_{v} = \frac{5}{2}

Plug in

x_{v} = \frac{5}{2}

to find the

y_{v}

value

y_{v} = - \frac{49}{4}

Therefore the parabola vertex is

(\frac{5}{2}, - \frac{49}{4})

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 1

Minimum (\frac{5}{2}, - \frac{49}{4})

- x^{2} - 2 x + 8

y = 4 x^{2} - 8

f (x) = 2 x^{2} - 12 x

y = (x + 2)^{2} - 3

f (x) = 2 x^{2} - 8 x - 5

What is the vertices y=x^2-5x-6 ?
The vertices y=x^2-5x-6 is Minimum (5/2 ,-49/4)