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

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

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

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

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

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

X Intercepts : (- 4, 0), (2, 0), Y Intercepts : (0, \frac{16}{3})

Asymptotes : Vertical x = 1, x = 3, Horizontal y = \frac{1}{2}

Extreme Points : Minimum (- 4, 0)

Interval Notation

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

Range : (- \infty, \infty)

Solution steps

Domain of

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

Range of

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

Axis interception points of

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

X Intercepts

: (- 4, 0), (2, 0),

Y Intercepts

: (0, \frac{16}{3})

Asymptotes of

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

Vertical

: x = 1, x = 3,

Horizontal

: y = \frac{1}{2}

Extreme Points of

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

Minimum

(- 4, 0)

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

f (x) = 3 + x + 3 - x

f (x) = x^{4} - x^{3} - x - 1

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

f (x) = 4 x^{3} - 6

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