What is the f(x)=log_{10}(((x^2-9))/((x-3))) ?

The f(x)=log_{10}(((x^2-9))/((x-3))) is

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f(x)=log_{10}(((x^2-9))/((x-3)))

f (x) = lo g_{10} (\frac{( x ^{2} - 9 )}{( x - 3 )})

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

Range : f (x) < lo g_{10} (6) or f (x) > lo g_{10} (6)

X Intercepts : (- 2, 0), Y Intercepts : (0, lo g_{10} (3))

Asymptotes : Vertical x = - 3

Extreme Points : None

Interval Notation

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

Range : (- \infty, lo g_{10} (6)) \cup (lo g_{10} (6), \infty)

Solution steps

Domain of

lo g_{10} (\frac{x ^{2} - 9}{x - 3}) : - 3 < x < 3 or x > 3

Range of

lo g_{10} (\frac{x ^{2} - 9}{x - 3}) : f (x) < lo g_{10} (6) or f (x) > lo g_{10} (6)

Axis interception points of

lo g_{10} (\frac{x ^{2} - 9}{x - 3}) :

X Intercepts

: (- 2, 0),

Y Intercepts

: (0, lo g_{10} (3))

Asymptotes of

lo g_{10} (\frac{x ^{2} - 9}{x - 3}) :

Vertical

: x = - 3

Extreme Points of

lo g_{10} (\frac{x ^{2} - 9}{x - 3}) :

None

f (x) = 5 x^{2} - 16 x + 1

y = \frac{( x ^{3} + 2 )}{3}

y = m (x - 3)

y = \frac{( m + 2 )}{( m - 3 ) \cdot x - 5 m + 7}

y = (3 e^{x} + x) \cdot cos (x)

What is the f(x)=log_{10}(((x^2-9))/((x-3))) ?
The f(x)=log_{10}(((x^2-9))/((x-3))) is