asymptotes of f(x)=(4x+1)/(x-2) — step-by-step solution

1) Let's find the asymptotes of

\frac{4 x + 1}{x - 2}

, 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 = 0

⇥ⓘ We add

2

to both sides

⇥Solution found:

x = 2

x

approaches

2

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 = 2

is therefore a vertical asymptote

⇥1.2) Let's start the euclidean division of

(4 x + 1)

(x - 2)

⇥

Expression in progress	Divide leading terms	Quotient	Quotient × divisor	Remainder after subtracting
$4 x + 1$	$\frac{4 x}{x}$	$4$	$4 \cdot (x - 2) = 4 x - 8$	$(4 x + 1) - (4 x - 8) = 9$

⇥So the result of the euclidean division is:

4

The quotient of the division is the constant

4

: the line to test is

y = 4

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

\frac{4 x + 1}{x - 2}

at both infinities

x

tends to +∞,

\frac{4 x + 1}{x - 2}

tends to

4

; as

x

tends to -∞,

\frac{4 x + 1}{x - 2}

tends to

4

The horizontal asymptote is

y = 4

; there is no slant asymptote

▶Asymptotes:

{x = 2, y = 4}

🖼/draw?f1=%284*x+%2B+1%29%2F%28x+-+2%29&v=x&x2=2&f3=4&b=-8.0&e=12.0&min=-1.31&max=8.41&dash=2,3