What is the f(x)=((4x^2-1/2 x))/((9x+8)) ?

The f(x)=((4x^2-1/2 x))/((9x+8)) is

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f(x)=((4x^2-1/2 x))/((9x+8))

f (x) = \frac{( 4 x ^{2} - \frac{1}{2} x )}{( 9 x + 8 )}

Domain : x < - \frac{8}{9} or x > - \frac{8}{9}

Range : f (x) \leq - \frac{8 73}{81} - \frac{137}{162} or f (x) \geq \frac{8 73}{81} - \frac{137}{162}

X Intercepts : (\frac{1}{8}, 0), (0, 0), Y Intercepts : (0, 0)

Asymptotes : Vertical x = - \frac{8}{9}, Slant y = \frac{4}{9} x - \frac{73}{162}

Extreme Points : Maximum (- \frac{8 + 73}{9}, - \frac{1168 + 137 73}{162 73}), Minimum (\frac{- 8 + 73}{9}, \frac{1168 - 137 73}{162 73})

Interval Notation

Domain : (- \infty, - \frac{8}{9}) \cup (- \frac{8}{9}, \infty)

Range : (- \infty, - \frac{8 73}{81} - \frac{137}{162}] \cup [\frac{8 73}{81} - \frac{137}{162}, \infty)

Solution steps

Domain of

\frac{4 x ^{2} - \frac{1}{2} x}{9 x + 8} : x < - \frac{8}{9} or x > - \frac{8}{9}

Range of

\frac{4 x ^{2} - \frac{1}{2} x}{9 x + 8} : f (x) \leq - \frac{8 73}{81} - \frac{137}{162} or f (x) \geq \frac{8 73}{81} - \frac{137}{162}

Axis interception points of

\frac{4 x ^{2} - \frac{1}{2} x}{9 x + 8} :

X Intercepts

: (\frac{1}{8}, 0), (0, 0),

Y Intercepts

: (0, 0)

Asymptotes of

\frac{4 x ^{2} - \frac{1}{2} x}{9 x + 8} :

Vertical

: x = - \frac{8}{9},

Slant

: y = \frac{4}{9} x - \frac{73}{162}

Extreme Points of

\frac{4 x ^{2} - \frac{1}{2} x}{9 x + 8} :

Maximum

(- \frac{8 + 73}{9}, - \frac{1168 + 137 73}{162 73}),

Minimum

(\frac{- 8 + 73}{9}, \frac{1168 - 137 73}{162 73})

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