What is the vertices-x^2+4x-6 ?

The vertices-x^2+4x-6 is Maximum (2,-2)

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vertices-x^2+4x-6

vertices

- x^{2} + 4 x - 6

Maximum (2, - 2)

Solution steps

y = - x^{2} + 4 x - 6

The parabola parameters are:

a = - 1, b = 4, c = - 6

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

x_{v} = - \frac{4}{2 ( - 1 )}

Simplify

- \frac{4}{2 ( - 1 )} : 2

x_{v} = 2

Plug in

x_{v} = 2

to find the

y_{v}

value

y_{v} = - 2

Therefore the parabola vertex is

(2, - 2)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = - 1

Maximum (2, - 2)

y = \frac{6 x}{7} \cdot \frac{4 x}{3}

y^{2} - 6 y - 4 x + 5 = 0

5 x (2 x + 4) - 3 x

f (x) = - 4 x^{2} - 12 x + 5

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