What is the vertices f(x)=x^2-5x-6 ?

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

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vertices f(x)=x^2-5x-6

vertices

f (x) = 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})

5 x^{2} + 7 y = 3

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

f (x) = 2 x^{2} - 80 x + 900

y = 5 x^{2}

f (x) = - (x + 3)^{2} + 5

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