What is the vertices y=2x^2+16x+29 ?

The vertices y=2x^2+16x+29 is Minimum (-4,-3)

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vertices y=2x^2+16x+29

vertices

y = 2 x^{2} + 16 x + 29

Minimum (- 4, - 3)

Solution steps

y = 2 x^{2} + 16 x + 29

The parabola parameters are:

a = 2, b = 16, c = 29

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

x_{v} = - \frac{16}{2 \cdot 2}

Simplify

x_{v} = - 4

Plug in

x_{v} = - 4

to find the

y_{v}

value

y_{v} = - 3

Therefore the parabola vertex is

(- 4, - 3)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 2

Minimum (- 4, - 3)

f (x) = - (x^{2} + 3 x - 4)

y = 4 x^{2} - 16

f (x) = x^{2} - 8 x + 17

y = 24000 x^{2} + 900 x - 150

16 - (12 x + 8)^{2}

What is the vertices y=2x^2+16x+29 ?
The vertices y=2x^2+16x+29 is Minimum (-4,-3)