asymptotes of f(x)=(5x)/(2x+3) — step-by-step solution

1) Let's find the asymptotes of

\frac{5 x}{2 x + 3}

, 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

2 x + 3 = 0

⇥ⓘ We subtract

3

from both sides

⇥

2 x = -3

⇥ⓘ Both sides are divided by

2

⇥Solution found:

x = - \frac{3}{2}

x

approaches

- \frac{3}{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 = - \frac{3}{2}

is therefore a vertical asymptote

⇥1.2) Let's start the euclidean division of

(5 x)

(2 x + 3)

⇥

Expression in progress	Divide leading terms	Quotient	Quotient × divisor	Remainder after subtracting
$5 x$	$\frac{5 x}{2 x}$	$\frac{5}{2}$	$\frac{5}{2} \cdot (2 x + 3) = 5 x + \frac{15}{2}$	$5 x - (5 x + \frac{15}{2}) = - \frac{15}{2}$

⇥So the result of the euclidean division is:

\frac{5}{2}

The quotient of the division is the constant

\frac{5}{2}

: the line to test is

y = \frac{5}{2}

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

\frac{5 x}{2 x + 3}

at both infinities

x

tends to +∞,

\frac{5 x}{2 x + 3}

tends to

\frac{5}{2}

; as

x

tends to -∞,

\frac{5 x}{2 x + 3}

tends to

\frac{5}{2}

The horizontal asymptote is

y = \frac{5}{2}

; there is no slant asymptote

▶Asymptotes:

{x = - \frac{3}{2}, y = \frac{5}{2}}

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