asymptotes of (5x+25)/(2x+7) — step-by-step solution

1) Let's find the asymptotes of

\frac{5 x + 25}{2 x + 7}

, 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 + 7 = 0

⇥ⓘ We subtract

7

from both sides

⇥

2 x = -7

⇥ⓘ Both sides are divided by

2

⇥Solution found:

x = - \frac{7}{2}

x

approaches

- \frac{7}{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{7}{2}

is therefore a vertical asymptote

⇥1.2) Perform polynomial long division of

(5 x + 25)

(2 x + 7)

⇥

Current expression	Divide leading terms	Quotient	Quotient × divisor	Remainder after subtracting
$5 x + 25$	$\frac{5 x}{2 x}$	$\frac{5}{2}$	$\frac{5}{2} \cdot (2 x + 7) = 5 x + \frac{35}{2}$	$(5 x + 25) - (5 x + \frac{35}{2}) = \frac{15}{2}$

⇥Therefore, 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 + 25}{2 x + 7}

at both infinities

x

tends to +∞,

\frac{5 x + 25}{2 x + 7}

tends to

\frac{5}{2}

; as

x

tends to -∞,

\frac{5 x + 25}{2 x + 7}

tends to

\frac{5}{2}

The horizontal asymptote is

y = \frac{5}{2}

; there is no slant asymptote

▶Asymptotes:

{x = - \frac{7}{2}, y = \frac{5}{2}}

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