asymptotes of f(x)=(3x)/(x^2-8) — step-by-step solution

Let's find the asymptotes of

\frac{3 x}{x^{2} - 8}

, 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

Let's solve in

ℝ

x^{2} - 8 = 0

The two numbers whose square is

8

are

- 2 \cdot \sqrt{2}

and

2 \cdot \sqrt{2}

Solutions found:

x = {- 2 \cdot \sqrt{2}, 2 \cdot \sqrt{2}}

x

approaches

- 2 \cdot \sqrt{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 \cdot \sqrt{2}

is therefore a vertical asymptote

x

approaches

2 \cdot \sqrt{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 \cdot \sqrt{2}

is therefore a vertical asymptote

At infinity, the numerator degree is smaller than the denominator degree: the line to test is

y = 0

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

\frac{3 x}{x^{2} - 8}

at both infinities

x

tends to +∞,

\frac{3 x}{x^{2} - 8}

tends to

0

; as

x

tends to -∞,

\frac{3 x}{x^{2} - 8}

tends to

0

The horizontal asymptote is

y = 0

; there is no slant asymptote

▶Asymptotes:

{x = - 2 \cdot \sqrt{2}, x = 2 \cdot \sqrt{2}, y = 0}

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