asymptotes of f(x)=(x+1)/(x-5) — step-by-step solution

1) Let's find the asymptotes of

\frac{x + 1}{x - 5}

, 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 - 5 = 0

⇥ⓘ We add

5

to both sides

⇥Solution found:

x = 5

x

approaches

5

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 = 5

is therefore a vertical asymptote

⇥1.2) Let's start the euclidean division of

(x + 1)

(x - 5)

⇥

Current expression	Divide leading terms	Quotient	Quotient × divisor	Remainder after subtracting
$x + 1$	$\frac{x}{x}$	$1$	$x - 5 = x - 5$	$(x + 1) - (x - 5) = 6$

⇥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 + 1}{x - 5}

at both infinities

x

tends to +∞,

\frac{x + 1}{x - 5}

tends to

1

; as

x

tends to -∞,

\frac{x + 1}{x - 5}

tends to

1

The horizontal asymptote is

y = 1

; there is no slant asymptote

▶Asymptotes:

{x = 5, y = 1}

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