What is the vertices f(x)=2(x-1)^2+4 ?

The vertices f(x)=2(x-1)^2+4 is Minimum (1,4)

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vertices f(x)=2(x-1)^2+4

vertices

f (x) = 2 (x - 1)^{2} + 4

Minimum (1, 4)

Solution steps

y = 2 (x - 1)^{2} + 4

Rewrite

y = 2 (x - 1)^{2} + 4

in the form

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

y = 2 x^{2} - 4 x + 6

The parabola parameters are:

a = 2, b = - 4, c = 6

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

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

Simplify

x_{v} = 1

Plug in

x_{v} = 1

to find the

y_{v}

value

y_{v} = 4

Therefore the parabola vertex is

(1, 4)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 2

Minimum (1, 4)

5 x^{2} + 20 x - 17

y = 2 x^{2} + 24 x - 6

- x^{2} + 4

y^{2} - 6 y = 8 x - 1

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

What is the vertices f(x)=2(x-1)^2+4 ?
The vertices f(x)=2(x-1)^2+4 is Minimum (1,4)