What is the f(x)=-1/2 (x+1)+2/3 (12x^2-6x-4) ?

The f(x)=-1/2 (x+1)+2/3 (12x^2-6x-4) is

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f(x)=-1/2 (x+1)+2/3 (12x^2-6x-4)

f (x) = - \frac{1}{2} (x + 1) + \frac{2}{3} (12 x^{2} - 6 x - 4)

Domain : - \infty < x < \infty

Range : f (x) \geq - \frac{1459}{384}

X Intercepts : (\frac{27 + 4377}{96}, 0), (\frac{27 - 4377}{96}, 0), Y Intercepts : (0, - \frac{19}{6})

Vertex : Minimum (\frac{9}{32}, - \frac{1459}{384})

Inverse : \frac{27 + 1152 x + 4377}{96}, \frac{27 - 1152 x + 4377}{96}

Interval Notation

Domain : (- \infty, \infty)

Range : [- \frac{1459}{384}, \infty)

Solution steps

Domain of

- \frac{1}{2} (x + 1) + \frac{2}{3} (12 x^{2} - 6 x - 4) : - \infty < x < \infty

Range of

- \frac{1}{2} (x + 1) + \frac{2}{3} (12 x^{2} - 6 x - 4) : f (x) \geq - \frac{1459}{384}

Axis interception points of

- \frac{1}{2} (x + 1) + \frac{2}{3} (12 x^{2} - 6 x - 4) :

X Intercepts

: (\frac{27 + 4377}{96}, 0), (\frac{27 - 4377}{96}, 0),

Y Intercepts

: (0, - \frac{19}{6})

Vertex of

- \frac{1}{2} (x + 1) + \frac{2}{3} (12 x^{2} - 6 x - 4) :

Minimum

(\frac{9}{32}, - \frac{1459}{384})

Inverse of

- \frac{1}{2} (x + 1) + \frac{2}{3} (12 x^{2} - 6 x - 4) : \frac{27 + 1152 x + 4377}{96}, \frac{27 - 1152 x + 4377}{96}

g (x) = (x + 3)^{2}

f (t) = sin^{4} (t)

y = 3 (x + 6)^{2} - 2

f (x) = 3 x^{2} - 12 x + 6

f (θ) = csc^{2} (θ) - cos^{2} (θ) cot^{2} (θ)

What is the f(x)=-1/2 (x+1)+2/3 (12x^2-6x-4) ?
The f(x)=-1/2 (x+1)+2/3 (12x^2-6x-4) is