asymptotes of f(x)=(2x^2-5x+7)/(x-2)

1) Let's find the asymptotes of

\frac{2 x^{2} - 5 x + 7}{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) Compute the quotient and remainder of

(2 x^{2} - 5 x + 7)

(x - 2)

⇥

Current expression	Divide leading terms	Quotient	Quotient × divisor	Remainder after subtracting
$2 x^{2} - 5 x + 7$	$\frac{2 x^{2}}{x}$	$2 x$	$2 x \cdot (x - 2) = 2 x^{2} - 4 x$	$(2 x^{2} - 5 x + 7) - (2 x^{2} - 4 x) = - x + 7$
$- x + 7$	$- \frac{x}{x}$	$-1$	$-1 \cdot (x - 2) = - x + 2$	$(- x + 7) - (- x + 2) = 5$

⇥Therefore, the result of the euclidean division is:

(2 x - 1)

The division lets us write

\frac{2 x^{2} - 5 x + 7}{x - 2} = (2 x - 1) + \frac{5}{x - 2}

: the line to test is

y = 2 x - 1

For a slant asymptote, the difference between the function and the line must tend to 0: check the limit of

\frac{5}{x - 2}

at both infinities

The difference tends to

0

at +∞ and to

0

at -∞: the curve therefore approaches the line at both infinities

The slant asymptote is

y = 2 x - 1

; there is no horizontal asymptote

▶Asymptotes:

{x = 2, y = 2 x - 1}

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