How do you find vertical, horizontal and oblique asymptotes for #(2x-2) /( 2x+2 )#?

Question

How do you find vertical, horizontal and oblique asymptotes for #(2x-2) /( 2x+2  )#?

Benjamin Achtenberg · Accepted Answer

vertical asymptote x = -1
horizontal asymptote y = 1

Begin by factorising numerator/denominator #rArr( cancel(2) (x-1))/(cancel(2) (x+1))=(x-1)/(x+1)# Vertical asymptotes occur as the denominator of a rational function tends to zero. To find the equation set the denominator equal to zero. solve: x + 1 = 0 → x = -1 is the asymptote Horizontal asymptotes occur as # lim_(x to +- oo) , f(x) to 0 # divide terms on numerator/denominator by x #(x/x-1/x)/(x/x+1/x)=(1-1/x)/(1+1/x)# as #x to +- oo , f(x) to (1-0)/(1+0)# #rArry=1" is the asymptote "# Oblique asymptotes occur when the degree of the numerator > degree of the denominator. This is not the case here (both of degree 1) hence there are no oblique asymptotes. graph{(x-1)/(x+1) [-10, 10, -5, 5]}

Theodore Amidon · Answer

undefined To find the vertical, horizontal, and oblique asymptotes for the function $ \frac{2x - 2}{2x + 2} $, follow these steps:

1. **Vertical Asymptotes:** Set the denominator equal to zero and solve for $ x $. If the denominator equals zero at a certain value of $ x $, then there is a vertical asymptote at that value.

$$ 2x + 2 = 0 $$
$$ x = -1 $$

So, there is a vertical asymptote at $ x = -1 $.

2. **Horizontal Asymptote:** Compare the degrees of the numerator and denominator. If the degree of the numerator is less than the degree of the denominator, the horizontal asymptote is at $ y = 0 $. If the degrees are equal, divide the leading coefficients to find the horizontal asymptote. If the degree of the numerator is greater, there is no horizontal asymptote.

In this case, both the numerator and denominator have a degree of 1, so we divide the leading coefficients:

$$ \frac{2}{2} = 1 $$

So, the horizontal asymptote is $ y = 1 $.

3. **Oblique Asymptote:** If the degree of the numerator is exactly one greater than the degree of the denominator, there is an oblique (slant) asymptote. To find it, perform polynomial long division or use the method of partial fractions to divide the numerator by the denominator. The quotient represents the oblique asymptote.

In this case, the degrees of the numerator and denominator are equal, so there is no oblique asymptote.

To summarize:
- Vertical asymptote: $ x = -1 $
- Horizontal asymptote: $ y = 1 $
- There is no oblique asymptote.