What is the vertices f(x)=-3x^2-18x+2 ?

The vertices f(x)=-3x^2-18x+2 is Maximum (-3,29)

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vertices f(x)=-3x^2-18x+2

vertices

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

Maximum (- 3, 29)

Solution steps

y = - 3 x^{2} - 18 x + 2

The parabola parameters are:

a = - 3, b = - 18, c = 2

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

x_{v} = - \frac{( - 18 )}{2 ( - 3 )}

Simplify

x_{v} = - 3

Plug in

x_{v} = - 3

to find the

y_{v}

value

y_{v} = 29

Therefore the parabola vertex is

(- 3, 29)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = - 3

Maximum (- 3, 29)

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

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

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

f (x) = - 2 x^{2} - 3 x + 1

f (x) = x^{2} - 4 x + 6

What is the vertices f(x)=-3x^2-18x+2 ?
The vertices f(x)=-3x^2-18x+2 is Maximum (-3,29)