asymptotes of f(x)=(x^2+9)/(x^2-9)

1) Let's find the asymptotes of

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

⇥1.1) Let's solve in

ℝ

x^{2} - 9 = 0

⇥The two numbers whose square is

9

are

-3

and

3

⇥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

⇥1.2) Compute the quotient and remainder of

(x^{2} + 9)

(x^{2} - 9)

⇥

Expression in progress	Divide leading terms	Quotient	Quotient × divisor	Remainder after subtracting
$x^{2} + 9$	$\frac{x^{2}}{x^{2}}$	$1$	$x^{2} - 9 = x^{2} - 9$	$(x^{2} + 9) - (x^{2} - 9) = 18$

⇥So the result of the euclidean division is:

1

The quotient of the division is the constant

1

: the line to test is

y = 1

For a horizontal asymptote, the function must tend to a finite value: check the limit of

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

at both infinities

x

tends to +∞,

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

tends to

1

; as

x

tends to -∞,

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

tends to

1

The horizontal asymptote is

y = 1

; there is no slant asymptote

▶Asymptotes:

{x = -3, x = 3, y = 1}

🖼/draw?f1=%28x%5E2+%2B+9%29%2F%28x%5E2+-+9%29&v=x&x2=-3&x3=3&f4=1&b=-10.0&e=10.0&min=-3.12&max=3.12&dash=2,3,4