What is the vertices f(x)=2x^2+8x+9 ?

The vertices f(x)=2x^2+8x+9 is Minimum (-2,1)

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vertices f(x)=2x^2+8x+9

vertices

f (x) = 2 x^{2} + 8 x + 9

Minimum (- 2, 1)

Solution steps

y = 2 x^{2} + 8 x + 9

The parabola parameters are:

a = 2, b = 8, c = 9

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} = 1

Therefore the parabola vertex is

(- 2, 1)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 2

Minimum (- 2, 1)

y = - x^{2} + 8 x - 15

f (x) = - 2 x^{2} + 8 x - 5

f (x) = - 2 x^{2} + 8 x - 9

f (x) = - 5 x^{2} + 10 x - 6

y = x^{2} - 4 x - 1

What is the vertices f(x)=2x^2+8x+9 ?
The vertices f(x)=2x^2+8x+9 is Minimum (-2,1)