What is the vertices f(x)=2x^2+8x-3 ?

The vertices f(x)=2x^2+8x-3 is Minimum (-2,-11)

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vertices f(x)=2x^2+8x-3

vertices

f (x) = 2 x^{2} + 8 x - 3

Minimum (- 2, - 11)

Solution steps

y = 2 x^{2} + 8 x - 3

The parabola parameters are:

a = 2, b = 8, c = - 3

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} = - 11

Therefore the parabola vertex is

(- 2, - 11)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 2

Minimum (- 2, - 11)

y^{2} = 20 x

y = 3 x^{2}

- 12 x^{2} - 29 x + 230

f (x) = 3 x^{2} - 6 x + 5

2 x^{2} + 4 x - 6

What is the vertices f(x)=2x^2+8x-3 ?
The vertices f(x)=2x^2+8x-3 is Minimum (-2,-11)