What is the vertices f(x)=2(x+14)^2-6 ?

The vertices f(x)=2(x+14)^2-6 is Minimum (-14,-6)

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vertices f(x)=2(x+14)^2-6

vertices

f (x) = 2 (x + 14)^{2} - 6

Minimum (- 14, - 6)

Solution steps

y = 2 (x + 14)^{2} - 6

Rewrite

y = 2 (x + 14)^{2} - 6

in the form

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

y = 2 x^{2} + 56 x + 386

The parabola parameters are:

a = 2, b = 56, c = 386

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

x_{v} = - \frac{56}{2 \cdot 2}

Simplify

x_{v} = - 14

Plug in

x_{v} = - 14

to find the

y_{v}

value

y_{v} = - 6

Therefore the parabola vertex is

(- 14, - 6)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 2

Minimum (- 14, - 6)

y = 3 x^{2} - 6 x + 5

x^{2} - 2

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

y = (x + 4)^{2}

f (x) = (x + 1)^{2} - 3

What is the vertices f(x)=2(x+14)^2-6 ?
The vertices f(x)=2(x+14)^2-6 is Minimum (-14,-6)