asymptotes of f(x)=1/(-2x^2+2x+12)

Let's find the asymptotes of

\frac{1}{- 2 x^{2} + 2 x + 12}

, 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

- 2 x^{2} + 2 x + 12 = 0

Since

-2 \neq 0

, divide both sides by

-2

x^{2} - x - 6 = 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 = -1

and

c = -6

Δ = {(-1)}^{2} - 4 \cdot 1 \cdot (-6)

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

{(-1)}^{2}

(-1) \cdot (-1)

1

Δ = 1 - 4 \cdot (-6)

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

- 4 \cdot (-6)

24

Δ = 1 + 24

ⓘ Start from

1

and move forward

24

on the number line:

1 + 24

25

Δ = 25

Δ > 0

, the equation has two roots:

The quadratic formula gives

x_{1}

\frac{- b - \sqrt{Δ}}{2 a}

\frac{- (-1) - \sqrt{25}}{2 \cdot 1}

x_{1} = \frac{1 - \sqrt{25}}{2}

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

\sqrt{25}

5

;

5^{2}

25

x_{1} = \frac{1 - 5}{2}

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

1 - 5

-4

x_{1} = - \frac{4}{2}

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

2

, their greatest common divisor:

- \frac{4}{2}

-2

x_{1} = -2

The quadratic formula gives

x_{2}

\frac{- b + \sqrt{Δ}}{2 a}

\frac{- (-1) + \sqrt{25}}{2 \cdot 1}

x_{2} = \frac{1 + \sqrt{25}}{2}

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

\sqrt{25}

5

;

5^{2}

25

x_{2} = \frac{1 + 5}{2}

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

1 + 5

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 = {-2, 3}

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

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{1}{- 2 x^{2} + 2 x + 12}

at both infinities

x

tends to +∞,

\frac{1}{- 2 x^{2} + 2 x + 12}

tends to

0

; as

x

tends to -∞,

\frac{1}{- 2 x^{2} + 2 x + 12}

tends to

0

The horizontal asymptote is

y = 0

; there is no slant asymptote

▶Asymptotes:

{x = -2, x = 3, y = 0}

🖼/draw?f1=1%2F%28-2*x%5E2+%2B+2*x+%2B+12%29&v=x&x2=-2&x3=3&f4=0&b=-9.5&e=10.5&min=-0.10119047619047619&max=0.11309523809523808&dash=2,3,4