What is the vertices f(x)=x^2-8x-9 ?

The vertices f(x)=x^2-8x-9 is Minimum (4,-25)

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vertices f(x)=x^2-8x-9

vertices

f (x) = x^{2} - 8 x - 9

Minimum (4, - 25)

Solution steps

y = x^{2} - 8 x - 9

The parabola parameters are:

a = 1, b = - 8, c = - 9

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

x_{v} = - \frac{( - 8 )}{2 \cdot 1}

Simplify

x_{v} = 4

Plug in

x_{v} = 4

to find the

y_{v}

value

y_{v} = - 25

Therefore the parabola vertex is

(4, - 25)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 1

Minimum (4, - 25)

X^{2} + 2 X - 8

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

f (x) = 2 x^{2} + 6 x + 1

x^{2} + 10 x + 11

y = x^{2} + 2 x - 35

What is the vertices f(x)=x^2-8x-9 ?
The vertices f(x)=x^2-8x-9 is Minimum (4,-25)