What is the vertices f(x)=x^2-6x-7 ?

The vertices f(x)=x^2-6x-7 is Minimum (3,-16)

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vertices f(x)=x^2-6x-7

vertices

f (x) = x^{2} - 6 x - 7

Minimum (3, - 16)

Solution steps

y = x^{2} - 6 x - 7

The parabola parameters are:

a = 1, b = - 6, c = - 7

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

x_{v} = - \frac{( - 6 )}{2 \cdot 1}

Simplify

x_{v} = 3

Plug in

x_{v} = 3

to find the

y_{v}

value

y_{v} = - 16

Therefore the parabola vertex is

(3, - 16)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 1

Minimum (3, - 16)

x^{2} - 2 x - 8

x^{2} - 6 x + 10

f (x) = x^{2} + 4 x - 3

f (x) = x^{2} + 10 x + 9

y = - 3 x^{2} - 12 x - 9

What is the vertices f(x)=x^2-6x-7 ?
The vertices f(x)=x^2-6x-7 is Minimum (3,-16)