What is the f(v)=((-3v-2))/((3v^2+4v+3)) ?

The f(v)=((-3v-2))/((3v^2+4v+3)) is

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f(v)=((-3v-2))/((3v^2+4v+3))

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

Domain : - \infty < v < \infty

Range : - \frac{3}{2 5} \leq f (v) \leq \frac{3}{2 5}

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

Asymptotes : Horizontal y = 0

Extreme Points : Maximum (- \frac{2 + 5}{3}, \frac{3}{2 5}), Minimum (\frac{- 2 + 5}{3}, - \frac{3}{2 5})

Interval Notation

Domain : (- \infty, \infty)

Range : [- \frac{3}{2 5}, \frac{3}{2 5}]

Solution steps

Domain of

\frac{- 3 v - 2}{3 v ^{2} + 4 v + 3} : - \infty < v < \infty

Range of

\frac{- 3 v - 2}{3 v ^{2} + 4 v + 3} : - \frac{3}{2 5} \leq f (v) \leq \frac{3}{2 5}

Axis interception points of

\frac{- 3 v - 2}{3 v ^{2} + 4 v + 3} :

X Intercepts

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

Y Intercepts

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

Asymptotes of

\frac{- 3 v - 2}{3 v ^{2} + 4 v + 3} :

Horizontal

: y = 0

Extreme Points of

\frac{- 3 v - 2}{3 v ^{2} + 4 v + 3} :

Maximum

(- \frac{2 + 5}{3}, \frac{3}{2 5}),

Minimum

(\frac{- 2 + 5}{3}, - \frac{3}{2 5})

f (t) = \frac{( t ^{n - 1} )}{( 1 - e ^{- t} )}

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

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

f (x) = - 2 x^{2} - 5 x

s (t) = 976 [(0.835) t - 1] + 176 t