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

The vertices f(x)=2x^2+8x+3 is Minimum (-2,-5)

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

vertices

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

Minimum (- 2, - 5)

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} = - 5

Therefore the parabola vertex is

(- 2, - 5)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 2

Minimum (- 2, - 5)

y = 2 x^{2} - 7

y = - 3 x^{2} - 36 x - 104

f (x) = x^{2} - 6 x + 11

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

f (x) = x^{2} + 10 x + 21

What is the vertices f(x)=2x^2+8x+3 ?
The vertices f(x)=2x^2+8x+3 is Minimum (-2,-5)