asymptotes of (x^2-2x-35)/(x^2-16)

1) Let's find the asymptotes of

\frac{x^{2} - 2 x - 35}{x^{2} - 16}

, 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} - 16 = 0

⇥The two numbers whose square is

16

are

-4

and

4

⇥Solutions obtained:

x = {-4, 4}

x

approaches

-4

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 = -4

is therefore a vertical asymptote

x

approaches

4

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 = 4

is therefore a vertical asymptote

⇥1.2) Perform polynomial long division of

(x^{2} - 2 x - 35)

(x^{2} - 16)

⇥

Current expression	Divide leading terms	Quotient	Quotient × divisor	Remainder after subtracting
$x^{2} - 2 x - 35$	$\frac{x^{2}}{x^{2}}$	$1$	$x^{2} - 16 = x^{2} - 16$	$(x^{2} - 2 x - 35) - (x^{2} - 16) = - 2 x - 19$

⇥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} - 2 x - 35}{x^{2} - 16}

at both infinities

x

tends to +∞,

\frac{x^{2} - 2 x - 35}{x^{2} - 16}

tends to

1

; as

x

tends to -∞,

\frac{x^{2} - 2 x - 35}{x^{2} - 16}

tends to

1

The horizontal asymptote is

y = 1

; there is no slant asymptote

▶Asymptotes:

{x = -4, x = 4, y = 1}

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