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

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

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vertices (x-4)^2+2

vertices

(x - 4)^{2} + 2

Minimum (4, 2)

Solution steps

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

Rewrite

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

in the form

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

y = x^{2} - 8 x + 18

The parabola parameters are:

a = 1, b = - 8, c = 18

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} = 2

Therefore the parabola vertex is

(4, 2)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 1

Minimum (4, 2)

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

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

y = 2 x^{2} + 15 x

f (x) = - x^{2} + 10 x - 16

f (x) = - x^{2} + 5 x - 6