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

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

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

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

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

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

X Intercepts : (1, 0), (4, 0), Y Intercepts : (0, - 2)

Asymptotes : Vertical x = 2, Slant y = x - 3

Extreme Points : None

Interval Notation

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

Range : (- \infty, \infty)

Solution steps

Domain of

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

Range of

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

Axis interception points of

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

X Intercepts

: (1, 0), (4, 0),

Y Intercepts

: (0, - 2)

Asymptotes of

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

Vertical

: x = 2,

Slant

: y = x - 3

Extreme Points of

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

None

f (x) = 8 - 3 x

f (x) = lo g_{3} (x + 2) + 1

r, r = \frac{3}{1 + cos ( θ )}

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

y = lo g_{3} (x) + 1

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