What is the f(x)=(x^2+4x-5)/((x^2-9)(x+1)) ?

The f(x)=(x^2+4x-5)/((x^2-9)(x+1)) is

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f(x)=(x^2+4x-5)/((x^2-9)(x+1))

f (x) = \frac{x ^{2} + 4 x - 5}{( x ^{2} - 9 ) ( x + 1 )}

Domain : x < - 3 or - 3 < x < - 1 or - 1 < x < 3 or x > 3

Range : - \infty < f (x) < \infty

X Intercepts : (1, 0), (- 5, 0), Y Intercepts : (0, \frac{5}{9})

Asymptotes : Vertical x = - 3, x = - 1, x = 3, Horizontal y = 0

Extreme Points : Minimum (- 8.21589 \dots, - 0.07020 \dots), Maximum (- 2.10866 \dots, - 1.78042 \dots)

Interval Notation

Domain : (- \infty, - 3) \cup (- 3, - 1) \cup (- 1, 3) \cup (3, \infty)

Range : (- \infty, \infty)

Solution steps

Domain of

\frac{x ^{2} + 4 x - 5}{( x ^{2} - 9 ) ( x + 1 )} : x < - 3 or - 3 < x < - 1 or - 1 < x < 3 or x > 3

Range of

\frac{x ^{2} + 4 x - 5}{( x ^{2} - 9 ) ( x + 1 )} : - \infty < f (x) < \infty

Axis interception points of

\frac{x ^{2} + 4 x - 5}{( x ^{2} - 9 ) ( x + 1 )} :

X Intercepts

: (1, 0), (- 5, 0),

Y Intercepts

: (0, \frac{5}{9})

Asymptotes of

\frac{x ^{2} + 4 x - 5}{( x ^{2} - 9 ) ( x + 1 )} :

Vertical

: x = - 3, x = - 1, x = 3,

Horizontal

: y = 0

Extreme Points of

\frac{x ^{2} + 4 x - 5}{( x ^{2} - 9 ) ( x + 1 )} :

Minimum

(- 8.21589 \dots, - 0.07020 \dots),

Maximum

(- 2.10866 \dots, - 1.78042 \dots)

f (x) = ln (8 - x)

y = x^{4} - 2 x^{2} + 5

f (x) = 8 - 4 x

y = \frac{1}{4 x + 2}

f (k) = k^{2} - 10 k + 24

What is the f(x)=(x^2+4x-5)/((x^2-9)(x+1)) ?
The f(x)=(x^2+4x-5)/((x^2-9)(x+1)) is