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

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

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

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

Domain : x < - 1 or - 1 < x < 1 or 1 < x \leq 2

Range : f (x) < \frac{1}{5 2} or f (x) > \frac{1}{5 2}

X Intercepts : (2, 0), Y Intercepts : (0, \frac{2}{5})

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

Extreme Points : Minimum (2, 0)

Interval Notation

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

Range : (- \infty, \frac{1}{5 2}) \cup (\frac{1}{5 2}, \infty)

Solution steps

Domain of

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

Range of

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

Axis interception points of

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

X Intercepts

: (2, 0),

Y Intercepts

: (0, \frac{2}{5})

Asymptotes of

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

Vertical

: x = - 1,

Horizontal

: y = 0

Extreme Points of

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

Minimum

(2, 0)

f (x) = \frac{x ^{3} - 27}{x - 3}

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

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

y = \frac{1}{2} x^{2} - 4

y = \frac{x}{x - 1}

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