What is the y=((2x^2+2)^2)/((x+1)^2(3x+2)) ?

The y=((2x^2+2)^2)/((x+1)^2(3x+2)) is

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y=((2x^2+2)^2)/((x+1)^2(3x+2))

y = \frac{( 2 x ^{2} + 2 ) ^{2}}{( x + 1 ) ^{2} ( 3 x + 2 )}

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

Range : f (x) \leq - 12.80044 \dots or f (x) \geq 0.74931 \dots

Y Intercepts : (0, 2)

Asymptotes : Vertical x = - 1, x = - \frac{2}{3}, Slant y = \frac{4}{3} x - \frac{32}{9}

Extreme Points : Maximum (- 4.28400 \dots, - 12.80044 \dots), Maximum (- 0.76308 \dots, - 616.83775 \dots), Minimum (0.71375 \dots, 0.74931 \dots)

Interval Notation

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

Range : (- \infty, - 12.80044 \dots] \cup [0.74931 \dots, \infty)

Solution steps

Domain of

\frac{( 2 x ^{2} + 2 ) ^{2}}{( x + 1 ) ^{2} ( 3 x + 2 )} : x < - 1 or - 1 < x < - \frac{2}{3} or x > - \frac{2}{3}

Range of

\frac{( 2 x ^{2} + 2 ) ^{2}}{( x + 1 ) ^{2} ( 3 x + 2 )} : f (x) \leq - 12.80044 \dots or f (x) \geq 0.74931 \dots

Axis interception points of

\frac{( 2 x ^{2} + 2 ) ^{2}}{( x + 1 ) ^{2} ( 3 x + 2 )} :

Y Intercepts

: (0, 2)

Asymptotes of

\frac{( 2 x ^{2} + 2 ) ^{2}}{( x + 1 ) ^{2} ( 3 x + 2 )} :

Vertical

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

Slant

: y = \frac{4}{3} x - \frac{32}{9}

Extreme Points of

\frac{( 2 x ^{2} + 2 ) ^{2}}{( x + 1 ) ^{2} ( 3 x + 2 )} :

Maximum

(- 4.28400 \dots, - 12.80044 \dots),

Maximum

(- 0.76308 \dots, - 616.83775 \dots),

Minimum

(0.71375 \dots, 0.74931 \dots)

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

f (x) = x^{3} - 3 x + 8

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

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

f (x) = x^{1.7}

What is the y=((2x^2+2)^2)/((x+1)^2(3x+2)) ?
The y=((2x^2+2)^2)/((x+1)^2(3x+2)) is