What is the vertices y=3x^2+9x+9 ?

The vertices y=3x^2+9x+9 is Minimum (-3/2 , 9/4)

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vertices y=3x^2+9x+9

vertices

y = 3 x^{2} + 9 x + 9

Minimum (- \frac{3}{2}, \frac{9}{4})

Solution steps

y = 3 x^{2} + 9 x + 9

The parabola parameters are:

a = 3, b = 9, c = 9

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

x_{v} = - \frac{9}{2 \cdot 3}

Simplify

- \frac{9}{2 \cdot 3} : - \frac{3}{2}

x_{v} = - \frac{3}{2}

Plug in

x_{v} = - \frac{3}{2}

to find the

y_{v}

value

y_{v} = \frac{9}{4}

Therefore the parabola vertex is

(- \frac{3}{2}, \frac{9}{4})

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 3

Minimum (- \frac{3}{2}, \frac{9}{4})

f (x) = x^{2} + 10 x + 7

f (x) = x^{2} - 3 x - 18

- 2 x + x (x - 2) - 2 x (x - 1) - (x - 3)

(5 x - 8)^{2}

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

What is the vertices y=3x^2+9x+9 ?
The vertices y=3x^2+9x+9 is Minimum (-3/2 , 9/4)