asymptotes of (x^3)/(x^2-9) — step-by-step solution

Let's find the asymptotes of

\frac{x^{3}}{x^{2} - 9}

, for real

x

Vertical asymptotes: find the excluded values, where the denominator is zero. Then check whether the curve goes to infinity near these values

⇥Let's solve in

ℝ

x^{2} - 9 = 0

⇥Solutions obtained:

x = {-3, 3}

x

approaches

-3

from the left (smaller values), the function tends to

-∞

; from the right (larger values), it tends to

+∞

The function tends to infinity on at least one side: the line

x = -3

is therefore a vertical asymptote

x

approaches

3

from the left (smaller values), the function tends to

-∞

; from the right (larger values), it tends to

+∞

The function tends to infinity on at least one side: the line

x = 3

is therefore a vertical asymptote

⇥Perform polynomial long division of

x^{3}

x^{2} - 9

⇥

Current expression	Divide leading terms	Quotient	Quotient × divisor	Remainder after subtracting
$x^{3}$	$\frac{x^{3}}{x^{2}}$	$x$	$x \cdot (x^{2} - 9) = x^{3} - 9 x$	$x^{3} - (x^{3} - 9 x) = 9 x$

⇥Therefore, the result of the euclidean division is:

x

The division lets us write

\frac{x^{3}}{x^{2} - 9} = x + \frac{9 x}{x^{2} - 9}

: the line to test is

y = x

For a slant asymptote, the difference between the function and the line must tend to 0: check the limit of

\frac{9 x}{x^{2} - 9}

at both infinities

The difference tends to

0

at +∞ and to

0

at -∞: the curve therefore approaches the line at both infinities

The slant asymptote is

y = x

; there is no horizontal asymptote

▶Asymptotes:

{x = -3, x = 3, y = x}

🖼/draw?f1=x%5E3%2F%28x%5E2+-+9%29&v=x&x2=-3&x3=3&f4=x&b=-10.0&e=10.0&min=-12.1375&max=12.0125&dash=2,3,4