asymptotes of f(x)=(x-2)/(x^2+2x-15)

Let's find the asymptotes of

\frac{x - 2}{x^{2} + 2 x - 15}

, 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} + 2 x - 15 = 0

Given the quadratic form

a x^{2} + b x + c

, let's compute the discriminant:

Δ = b^{2} - 4 a c

. Here

a = 1

b = 2

and

c = -15

Δ = 2^{2} - 4 \cdot 1 \cdot (-15)

ⓘ Powers are computed before multiplications, additions and subtractions. An integer power is a repeated product:

2^{2}

2 \cdot 2

4

Δ = 4 - 4 \cdot (-15)

ⓘ Multiplications are done before additions and subtractions. Two minus signs cancel out, the product of two negative numbers is positive:

- 4 \cdot (-15)

4 \cdot 15

4 \cdot (10 + 5)

4 \cdot 10 + 4 \cdot 5

40 + 20

60

Δ = 4 + 60

ⓘ Start from

4

and move forward

60

on the number line:

4 + 60

64

Δ = 64

Δ > 0

, the equation has two roots:

The quadratic formula gives

x_{1}

\frac{- b - \sqrt{Δ}}{2 a}

\frac{- 2 - \sqrt{64}}{2 \cdot 1}

x_{1} = \frac{- 2 - \sqrt{64}}{2}

ⓘ The square root of a number is the positive number whose square is that number:

\sqrt{64}

8

;

8^{2}

64

x_{1} = \frac{- 2 - 8}{2}

ⓘ The numerator and the denominator are computed before the division:

- 2 - 8

-10

x_{1} = - \frac{10}{2}

ⓘ We simplify the fraction by dividing the numerator and the denominator by

2

, their greatest common divisor:

- \frac{10}{2}

-5

x_{1} = -5

The quadratic formula gives

x_{2}

\frac{- b + \sqrt{Δ}}{2 a}

\frac{- 2 + \sqrt{64}}{2 \cdot 1}

x_{2} = \frac{- 2 + \sqrt{64}}{2}

ⓘ The square root of a number is the positive number whose square is that number:

\sqrt{64}

8

;

8^{2}

64

x_{2} = \frac{- 2 + 8}{2}

ⓘ The numerator and the denominator are computed before the division:

- 2 + 8

6

x_{2} = \frac{6}{2}

ⓘ We simplify the fraction by dividing the numerator and the denominator by

2

, their greatest common divisor:

\frac{6}{2}

3

x_{2} = 3

Solutions obtained:

x = {-5, 3}

x

approaches

-5

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 = -5

is therefore a vertical asymptote

x

approaches

3

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 = 3

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{x - 2}{x^{2} + 2 x - 15}

at both infinities

x

tends to +∞,

\frac{x - 2}{x^{2} + 2 x - 15}

tends to

0

; as

x

tends to -∞,

\frac{x - 2}{x^{2} + 2 x - 15}

tends to

0

The horizontal asymptote is

y = 0

; there is no slant asymptote

▶Asymptotes:

{x = -5, x = 3, y = 0}

🖼/draw?f1=%28x+-+2%29%2F%28x%5E2+%2B+2*x+-+15%29&v=x&x2=-5&x3=3&f4=0&b=-13.0&e=11.0&min=-0.4207815606644494&max=0.44862619488514105&dash=2,3,4