What is the vertices f(x)=(x-2)^2+4 ?

The vertices f(x)=(x-2)^2+4 is Minimum (2,4)

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vertices f(x)=(x-2)^2+4

vertices

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

Minimum (2, 4)

Solution steps

y = (x - 2)^{2} + 4

Rewrite

y = (x - 2)^{2} + 4

in the form

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

y = x^{2} - 4 x + 8

The parabola parameters are:

a = 1, b = - 4, c = 8

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} = 4

Therefore the parabola vertex is

(2, 4)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 1

Minimum (2, 4)

f (x) = (x - 2)^{2} - 1

f (x) = - 0.02 x^{2} + x + 6

y = x^{2} - 1

+ \frac{- 1}{2} x^{2} + 18

x^{2} - 4 x + 12

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