What is the vertices f(x)=-6x^2+17x-12 ?

The vertices f(x)=-6x^2+17x-12 is Maximum (17/12 , 1/24)

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vertices f(x)=-6x^2+17x-12

vertices

f (x) = - 6 x^{2} + 17 x - 12

Maximum (\frac{17}{12}, \frac{1}{24})

Solution steps

y = - 6 x^{2} + 17 x - 12

The parabola parameters are:

a = - 6, b = 17, c = - 12

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

x_{v} = - \frac{17}{2 ( - 6 )}

Simplify

x_{v} = \frac{17}{12}

Plug in

x_{v} = \frac{17}{12}

to find the

y_{v}

value

y_{v} = \frac{1}{24}

Therefore the parabola vertex is

(\frac{17}{12}, \frac{1}{24})

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = - 6

Maximum (\frac{17}{12}, \frac{1}{24})

f (x) = 4 x^{2} - 16 x + 10

2 x^{2} - 6 x + 4

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

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

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

What is the vertices f(x)=-6x^2+17x-12 ?
The vertices f(x)=-6x^2+17x-12 is Maximum (17/12 , 1/24)