What is the y=log_{10}(x^5)(2x^2+1) ?

The y=log_{10}(x^5)(2x^2+1) is Domain: x>0 X Intercepts: (1,0) Asymptotes: Vertical x=0

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y=log_{10}(x^5)(2x^2+1)

y = lo g_{10} (x^{5}) (2 x^{2} + 1)

Domain : x > 0

X Intercepts : (1, 0)

Asymptotes : Vertical x = 0

Solution steps

Domain of

lo g_{10} (x^{5}) (2 x^{2} + 1) : x > 0

Axis interception points of

lo g_{10} (x^{5}) (2 x^{2} + 1) :

X Intercepts

: (1, 0)

Asymptotes of

lo g_{10} (x^{5}) (2 x^{2} + 1) :

Vertical

: x = 0

f (x) = 3^{2 x + 3} - 243

f (k) = 2 k^{4} - k^{2} + 8 k^{3} + 1

f (x) = x^{2} - 42 x - 451

y = 2 x^{2} + e^{x} + 7

f (p) = 7 p^{2} - 27 p - 7

What is the y=log_{10}(x^5)(2x^2+1) ?
The y=log_{10}(x^5)(2x^2+1) is Domain: x>0<br/>X Intercepts: (1,0)<br/>Asymptotes: Vertical x=0