What is the f(x)=(-6x^2+2x-1)/(3x^2-40x+2) ?

The f(x)=(-6x^2+2x-1)/(3x^2-40x+2) is

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f(x)=(-6x^2+2x-1)/(3x^2-40x+2)

f (x) = \frac{- 6 x ^{2} + 2 x - 1}{3 x ^{2} - 40 x + 2}

Domain : x < \frac{20 - 394}{3} or \frac{20 - 394}{3} < x < \frac{20 + 394}{3} or x > \frac{20 + 394}{3}

Range : f (x) \leq \frac{- 9 105 - 25}{788} or f (x) \geq \frac{9 105 - 25}{788}

Y Intercepts : (0, - \frac{1}{2})

Asymptotes : Vertical x = \frac{20 - 394}{3}, x = \frac{20 + 394}{3}, Horizontal y = - 2

Extreme Points : Maximum (\frac{1 - 105}{26}, - \frac{25 + 9 105}{788}), Minimum (\frac{1 + 105}{26}, \frac{9 105 - 25}{788})

Interval Notation

Domain : (- \infty, \frac{20 - 394}{3}) \cup (\frac{20 - 394}{3}, \frac{20 + 394}{3}) \cup (\frac{20 + 394}{3}, \infty)

Range : (- \infty, \frac{- 9 105 - 25}{788}] \cup [\frac{9 105 - 25}{788}, \infty)

Solution steps

Domain of

\frac{- 6 x ^{2} + 2 x - 1}{3 x ^{2} - 40 x + 2} : x < \frac{20 - 394}{3} or \frac{20 - 394}{3} < x < \frac{20 + 394}{3} or x > \frac{20 + 394}{3}

Range of

\frac{- 6 x ^{2} + 2 x - 1}{3 x ^{2} - 40 x + 2} : f (x) \leq \frac{- 9 105 - 25}{788} or f (x) \geq \frac{9 105 - 25}{788}

Axis interception points of

\frac{- 6 x ^{2} + 2 x - 1}{3 x ^{2} - 40 x + 2} :

Y Intercepts

: (0, - \frac{1}{2})

Asymptotes of

\frac{- 6 x ^{2} + 2 x - 1}{3 x ^{2} - 40 x + 2} :

Vertical

: x = \frac{20 - 394}{3}, x = \frac{20 + 394}{3},

Horizontal

: y = - 2

Extreme Points of

\frac{- 6 x ^{2} + 2 x - 1}{3 x ^{2} - 40 x + 2} :

Maximum

(\frac{1 - 105}{26}, - \frac{25 + 9 105}{788}),

Minimum

(\frac{1 + 105}{26}, \frac{9 105 - 25}{788})

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f (x) = x^{2} - 3 x - 30

f (x) = 6 + x + 2

f (t) = 8 sinh^{2} (3 t)

f (x) = lo g_{10} (4) (2 - x)

What is the f(x)=(-6x^2+2x-1)/(3x^2-40x+2) ?
The f(x)=(-6x^2+2x-1)/(3x^2-40x+2) is