What is the asymptotes of f(x)=(9x^2+36x+41)/(3x+5) ?

The asymptotes of f(x)=(9x^2+36x+41)/(3x+5) is Vertical: x=-5/3 ,Slant: y=3x+7

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asymptotes of f(x)=(9x^2+36x+41)/(3x+5)

asymptotes

f (x) = \frac{9 x ^{2} + 36 x + 41}{3 x + 5}

Vertical : x = - \frac{5}{3}, Slant : y = 3 x + 7

Solution steps

Vertical asymptotes of

\frac{9 x ^{2} + 36 x + 41}{3 x + 5} : x = - \frac{5}{3}

Horizontal Asymptotes of

\frac{9 x ^{2} + 36 x + 41}{3 x + 5} :

None

Slant Asymptotes of

\frac{9 x ^{2} + 36 x + 41}{3 x + 5} : y = 3 x + 7

Vertical : x = - \frac{5}{3}, Slant : y = 3 x + 7

f (x) = 2 tan (π x)

f (x) = (x - 15)^{2}, x \leq 15

f (x) = 20 x^{3}

x^{2} - 2 x + 5

f (x) = \frac{3 x - 5}{x + 1}

What is the asymptotes of f(x)=(9x^2+36x+41)/(3x+5) ?
The asymptotes of f(x)=(9x^2+36x+41)/(3x+5) is Vertical: x=-5/3 ,Slant: y=3x+7