What is the y=(3x^2+17x+10)/(6x^2-11x+4) ?

The y=(3x^2+17x+10)/(6x^2-11x+4) is

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y=(3x^2+17x+10)/(6x^2-11x+4)

y = \frac{3 x ^{2} + 17 x + 10}{6 x ^{2} - 11 x + 4}

Domain : x < \frac{1}{2} or \frac{1}{2} < x < \frac{4}{3} or x > \frac{4}{3}

Range : f (x) \leq \frac{- 6 2926 - 331}{25} or f (x) \geq \frac{6 2926 - 331}{25}

X Intercepts : (- \frac{2}{3}, 0), (- 5, 0), Y Intercepts : (0, \frac{5}{2})

Asymptotes : Vertical x = \frac{1}{2}, x = \frac{4}{3}, Horizontal y = \frac{1}{2}

Extreme Points : Minimum (- \frac{16 + 2926}{45}, - \frac{217913850 - 3950100 2926}{16458750}), Maximum (\frac{2926 - 16}{45}, - \frac{217913850 + 3950100 2926}{16458750})

Interval Notation

Domain : (- \infty, \frac{1}{2}) \cup (\frac{1}{2}, \frac{4}{3}) \cup (\frac{4}{3}, \infty)

Range : (- \infty, \frac{- 6 2926 - 331}{25}] \cup [\frac{6 2926 - 331}{25}, \infty)

Solution steps

Domain of

\frac{3 x ^{2} + 17 x + 10}{6 x ^{2} - 11 x + 4} : x < \frac{1}{2} or \frac{1}{2} < x < \frac{4}{3} or x > \frac{4}{3}

Range of

\frac{3 x ^{2} + 17 x + 10}{6 x ^{2} - 11 x + 4} : f (x) \leq \frac{- 6 2926 - 331}{25} or f (x) \geq \frac{6 2926 - 331}{25}

Axis interception points of

\frac{3 x ^{2} + 17 x + 10}{6 x ^{2} - 11 x + 4} :

X Intercepts

: (- \frac{2}{3}, 0), (- 5, 0),

Y Intercepts

: (0, \frac{5}{2})

Asymptotes of

\frac{3 x ^{2} + 17 x + 10}{6 x ^{2} - 11 x + 4} :

Vertical

: x = \frac{1}{2}, x = \frac{4}{3},

Horizontal

: y = \frac{1}{2}

Extreme Points of

\frac{3 x ^{2} + 17 x + 10}{6 x ^{2} - 11 x + 4} :

Minimum

(- \frac{16 + 2926}{45}, - \frac{217913850 - 3950100 2926}{16458750}),

Maximum

(\frac{2926 - 16}{45}, - \frac{217913850 + 3950100 2926}{16458750})

y = 5 cos (3 x - \frac{π}{4})

p (x) = x^{3} + 10 x^{2} + 3 x - 54

y = tan^{4} (3) x^{2}

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

f (x) = \frac{x ^{2} + 1}{2 x}