vertical and horizontal asymptotes

Question

Find the verticle and horizontal asymptotes of function f(x)=3x-1/x+4

 

find the verticle and horizontal asymptotes of function g(x)=x+7/x2-4

 

find the domain and function and express answer in interval notation  f(x)=4+2-7x+3

Answer 1

For your first two equations, the vertical asymptotes come from the zeroes of the denominator, so just set the denominator equal to zero and solve.  Also, the vertical asymptotes tell you if there are any restrictions on the domain.  Here's an example:

y = [x² + 2x - 3] / [x² - 5x - 6]

x² – 5x – 6 = 0 <- set the denominator equal to 0
(x – 6)(x + 1) = 0
x = 6 or –1 <- this function has 2 vertical asymptotes and the domain is all values of x except 6 or -1

 

To find horizontal asymptotes, there are 3 categories:

(1) If the highest power of the numerator and denominator are the same, just divide the leading terms (e.g. y = x² / 4x² = 1/4)

(2) If the highest power is in the denominator, the horizontal asymptote is always y=0

(3) If the highest power is in the numerator, you don't have horizontal asymptotes, you have slant asymptotes.

Answer 2

1.Function: f(x) = (3x - 1) / (x + 4)

 

A. A vertical asymptote has the equation x = c

where c is that value of x that makes a factor

in the denominator = 0.

 

The denominator of the function given is (x + 4).

If  x + 4 = 0

        -4  = -4

     x      = -4

 

Thus the vertical asymptote is: x = -4

 

B.  A rational function R(x) = p(x)/q(x) where 

p(x) is a polynomial of degree n and q(x) is a

polynomial of degree m. 

 

If n < m then the horizontal asymptote is y=0

or the x-axis.

 

If n = m i.e both the numerator and the

denominator are of the same degree, then the 

horizontal asymptote is:

y = quotient of the coefficients of the highest

      degree terms.

 

In this example the numerator is 3x - 1. The highest

degree of x is 1 and its coefficient = 3.

 

The denominator is x + 4 where x has degree 1 and

coefficient = 1.

 

Since n = m, the quotient of the coefficients of x in

each case = 3/1 = 3 

=> The horizontal asymptote is y = 3.

 

 

                              ______________________

 

 

2. Function: g(x)= (x + 7) / (x² - 4)

   

A. The denominator is (x² - 4) which factors to give:

     (x + 2) (x - 2)

     If x + 2 = 0 then x = -2

     If x - 2 = 0 then x = 2

     Thus the vertical asymptotes are x = 2 andx = -2.

 

B.  The numerator  x + 7 has degree 1 and x has a coefficient of 1.

     The denominator  x² - 4 has degree 2 and x² has coefficient of 1.

 

     In this case since n< m i.e the degree of the numerator is less than

     the degree of the denominator then the horizontal asymptote is 

     y = 0 or thex-axis.

 

                             _______________________

 

 

3. I am assuming that this question is incorrectly written and

    a possible alternative to the question could  be:

   

    Find the domain of the function and express the answer in

    interval notation: f(x) = 4x² - 7x + 3.

 

   If so, then the domain of this function consists of all the possible values

   that x can have. One must check to see if there are any restrictions

   on x. In this case there are none.

   

   In interval notation the domain of the function is: ( - ∞, +∞ )