What is the p(x)=(x^2+2x-3)/((x-1)(x^2-4)) ?

The p(x)=(x^2+2x-3)/((x-1)(x^2-4)) is

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p(x)=(x^2+2x-3)/((x-1)(x^2-4))

p (x) = \frac{x ^{2} + 2 x - 3}{( x - 1 ) ( x ^{2} - 4 )}

Domain : x < - 2 or - 2 < x < 1 or 1 < x < 2 or x > 2

Range : f (x) \leq - \frac{5 + 3}{8} or f (x) \geq - \frac{3 - 5}{8}

X Intercepts : (- 3, 0), Y Intercepts : (0, - \frac{3}{4})

Asymptotes : Vertical x = - 2, x = 2, Horizontal y = 0

Extreme Points : Minimum (- 3 - 5, - \frac{3 - 5}{8}), Maximum (- 3 + 5, - \frac{5 + 3}{8})

Interval Notation

Domain : (- \infty, - 2) \cup (- 2, 1) \cup (1, 2) \cup (2, \infty)

Range : (- \infty, - \frac{5 + 3}{8}] \cup [- \frac{3 - 5}{8}, \infty)

Solution steps

Domain of

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

Range of

\frac{x ^{2} + 2 x - 3}{( x - 1 ) ( x ^{2} - 4 )} : f (x) \leq - \frac{5 + 3}{8} or f (x) \geq - \frac{3 - 5}{8}

Axis interception points of

\frac{x ^{2} + 2 x - 3}{( x - 1 ) ( x ^{2} - 4 )} :

X Intercepts

: (- 3, 0),

Y Intercepts

: (0, - \frac{3}{4})

Asymptotes of

\frac{x ^{2} + 2 x - 3}{( x - 1 ) ( x ^{2} - 4 )} :

Vertical

: x = - 2, x = 2,

Horizontal

: y = 0

Extreme Points of

\frac{x ^{2} + 2 x - 3}{( x - 1 ) ( x ^{2} - 4 )} :

Minimum

(- 3 - 5, - \frac{3 - 5}{8}),

Maximum

(- 3 + 5, - \frac{5 + 3}{8})

(x - p)^{2}

f (x) = 11 x + 2

f (x) = 2 + lo g_{10} (x)

y = a x + 9

y = \frac{2}{3} x + \frac{1}{3}

What is the p(x)=(x^2+2x-3)/((x-1)(x^2-4)) ?
The p(x)=(x^2+2x-3)/((x-1)(x^2-4)) is