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

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

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

vertices

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

Minimum (2, - 9)

Solution steps

y = (x - 2)^{2} - 9

Rewrite

y = (x - 2)^{2} - 9

in the form

y = a x^{2} + b x + c

y = x^{2} - 4 x - 5

The parabola parameters are:

a = 1, b = - 4, c = - 5

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

x_{v} = - \frac{( - 4 )}{2 \cdot 1}

Simplify

x_{v} = 2

Plug in

x_{v} = 2

to find the

y_{v}

value

y_{v} = - 9

Therefore the parabola vertex is

(2, - 9)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 1

Minimum (2, - 9)

(3 x - 4) (4 x - 3) - 2 x

4 x^{2}

y = x^{2} - 2

x^{2} + 2 x - 3

4 x^{2} - 24 x + 34

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