What is the vertices y=2x^{(2)}+8x+5 ?

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

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vertices y=2x^{(2)}+8x+5

vertices

y = 2 x^{(2)} + 8 x + 5

Minimum (- 2, - 3)

Solution steps

y = 2 x^{2} + 8 x + 5

The parabola parameters are:

a = 2, b = 8, c = 5

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} = - 3

Therefore the parabola vertex is

(- 2, - 3)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 2

Minimum (- 2, - 3)

y = 2 3 x - 5

f (x) = \frac{- 1 - 7 x}{x - 2}

(- 7.13) (- 14)

f (x) = 4 cos (\frac{1}{5} π x + π) - 3

(10, 3), (4, - 2)

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