What is the vertices y=2(x+1)^2-8 ?

The vertices y=2(x+1)^2-8 is Minimum (-1,-8)

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vertices y=2(x+1)^2-8

vertices

y = 2 (x + 1)^{2} - 8

Minimum (- 1, - 8)

Solution steps

y = 2 (x + 1)^{2} - 8

Rewrite

y = 2 (x + 1)^{2} - 8

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} = - 8

Therefore the parabola vertex is

(- 1, - 8)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 2

Minimum (- 1, - 8)

25 - x^{2}

f (x) = sin (x) + cos (x)

f (x) = x^{4} - 32 x^{2} + 256

f (x) = 7 x^{3} - 2

f (x) = 4 x^{2} - 4 x + 21

What is the vertices y=2(x+1)^2-8 ?
The vertices y=2(x+1)^2-8 is Minimum (-1,-8)