What is the vertices-4x^2+6x-1 ?

The vertices-4x^2+6x-1 is Maximum (3/4 , 5/4)

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vertices-4x^2+6x-1

vertices

- 4 x^{2} + 6 x - 1

Maximum (\frac{3}{4}, \frac{5}{4})

Solution steps

y = - 4 x^{2} + 6 x - 1

The parabola parameters are:

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

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

x_{v} = - \frac{6}{2 ( - 4 )}

Simplify

- \frac{6}{2 ( - 4 )} : \frac{3}{4}

x_{v} = \frac{3}{4}

Plug in

x_{v} = \frac{3}{4}

to find the

y_{v}

value

y_{v} = \frac{5}{4}

Therefore the parabola vertex is

(\frac{3}{4}, \frac{5}{4})

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = - 4

Maximum (\frac{3}{4}, \frac{5}{4})

f (x) = x^{2} - 8 x + 18

y^{2} - 4 x - 4 = 0

y = 3 x^{2} + 6 x - 1

(8 x - 23)^{2} - 10

y = (x + 1)^{2} + 1