What is the vertices y=2x^2+24x-6 ?

The vertices y=2x^2+24x-6 is Minimum (-6,-78)

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vertices y=2x^2+24x-6

vertices

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

Minimum (- 6, - 78)

Solution steps

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

The parabola parameters are:

a = 2, b = 24, c = - 6

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

x_{v} = - \frac{24}{2 \cdot 2}

Simplify

x_{v} = - 6

Plug in

x_{v} = - 6

to find the

y_{v}

value

y_{v} = - 78

Therefore the parabola vertex is

(- 6, - 78)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 2

Minimum (- 6, - 78)

f (x) = x + 2

f (x) = 5 - \frac{x + 1}{3}

y = 3 (x + 4)^{2} + 1

y = - 1, (8, - 4)

f (x) = 9 - x^{2}

What is the vertices y=2x^2+24x-6 ?
The vertices y=2x^2+24x-6 is Minimum (-6,-78)