asymptotes of (3x^2-18x+24)/(x^2-4x)

1) Let's find the asymptotes of

\frac{3 x^{2} - 18 x + 24}{x^{2} - 4 x}

, 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} - 4 x = 0

⇥You can factorize the expression:

x^{2} - 4 x

into:

x \cdot (x - 4)

⇥A product is zero if and only if one of its factors is zero: we solve each factor equal to 0

⇥⇥1.1.1) Let's solve in

ℝ

x = 0

⇥⇥Solution found:

x = 0

⇥⇥1.1.2) Let's solve in

ℝ

x - 4 = 0

⇥⇥Solution obtained:

x = 4

x

approaches

0

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 = 0

is therefore a vertical asymptote

x

approaches

4

from the left (smaller values), the function tends to

\frac{3}{2}

; from the right (larger values), it tends to

\frac{3}{2}

The limits at

4

are finite: this is a hole, not a vertical asymptote

⇥1.2) Perform polynomial long division of

(3 x^{2} - 18 x + 24)

(x^{2} - 4 x)

⇥

Current expression	Divide leading terms	Quotient	Quotient × divisor	Remainder after subtracting
$3 x^{2} - 18 x + 24$	$\frac{3 x^{2}}{x^{2}}$	$3$	$3 \cdot (x^{2} - 4 x) = 3 x^{2} - 12 x$	$(3 x^{2} - 18 x + 24) - (3 x^{2} - 12 x) = - 6 x + 24$

⇥Therefore, the result of the euclidean division is:

3

The quotient of the division is the constant

3

: the line to test is

y = 3

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

\frac{3 x^{2} - 18 x + 24}{x^{2} - 4 x}

at both infinities

x

tends to +∞,

\frac{3 x^{2} - 18 x + 24}{x^{2} - 4 x}

tends to

3

; as

x

tends to -∞,

\frac{3 x^{2} - 18 x + 24}{x^{2} - 4 x}

tends to

3

The horizontal asymptote is

y = 3

; there is no slant asymptote

▶Asymptotes:

{x = 0, y = 3}

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