What is the vertices f(x)=-x^2+5x-6 ?

The vertices f(x)=-x^2+5x-6 is Maximum (5/2 , 1/4)

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vertices f(x)=-x^2+5x-6

vertices

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

Maximum (\frac{5}{2}, \frac{1}{4})

Solution steps

y = - x^{2} + 5 x - 6

The parabola parameters are:

a = - 1, b = 5, c = - 6

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

x_{v} = - \frac{5}{2 ( - 1 )}

Simplify

x_{v} = \frac{5}{2}

Plug in

x_{v} = \frac{5}{2}

to find the

y_{v}

value

y_{v} = \frac{1}{4}

Therefore the parabola vertex is

(\frac{5}{2}, \frac{1}{4})

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = - 1

Maximum (\frac{5}{2}, \frac{1}{4})

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

y = 3 x^{2} - 3 x - 6

f (x) = x^{2} - 3 x + 10

- 2 x^{2} + 12 x - 20

y^{2} + 2 x + 6 y + 17 = 0

What is the vertices f(x)=-x^2+5x-6 ?
The vertices f(x)=-x^2+5x-6 is Maximum (5/2 , 1/4)