What is the vertices y=2x^2+8x-34 ?

The vertices y=2x^2+8x-34 is Minimum (-2,-42)

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vertices y=2x^2+8x-34

vertices

y = 2 x^{2} + 8 x - 34

Minimum (- 2, - 42)

Solution steps

y = 2 x^{2} + 8 x - 34

The parabola parameters are:

a = 2, b = 8, c = - 34

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

x_{v} = - \frac{8}{2 \cdot 2}

Simplify

x_{v} = - 2

Plug in

x_{v} = - 2

to find the

y_{v}

value

y_{v} = - 42

Therefore the parabola vertex is

(- 2, - 42)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 2

Minimum (- 2, - 42)

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

3 x^{2} + 2 x + 1

3 x^{2} + 8 y = 0

3 x^{2} + 2 x - 3

2 x^{2} + 4 x + 1

What is the vertices y=2x^2+8x-34 ?
The vertices y=2x^2+8x-34 is Minimum (-2,-42)