Rational Functions

-- Section 5.6 --
 
Find the vertical and horizontal asymptotes, if any.

Solve

            Step

   

 VA: Set the denominator equal to zero.

 Set x - 3 = 0   So, x  = 3. Obviously, f(3) = undefined.  What happens as x approaches 3?  i.e. x → 3

From the left hand side of 3

From the right hand side of 3

The Vertical Asymptote, VA, is the line, x  = 3.

From the sketch of the graph we see y = 0 is the horizontal asymptote.  Look what happens when we plug in large and small numbers.

The Horizontal Asymptote, HA, is the line, y  = 0

Whenever the degree of the numerator is less than the degree of the denominator, the HA is y = 0.

As x →∞ (or -∞) , f (x) → 0 because the denominator will increase much faster than the numerator.

  
 Find the Vertical Asymptotes, if any.

[Solution]

Find the vertical and horizontal asymptotes, if any.  Then find x and y intercepts, if any.
 

Solve

            Step

    

Vertical Asymptote:  x  - 2  = 0, x  = 2

 The vertical asymptote is the line x = 2

Horizontal Asymptote:  y = 3

 The horizontal asymptote is the line y = 3  

If the degree of the numerator and the denominator are the same, as x approaches ±∞ , then y approaches the coefficients of the of the leading term in  the numerator and denominator.

  x - intercept: (-4/3, 0)

 Set y = g(x) = 0. The only way a fraction can equal zero is if the numerator is zero.

 

y - intercept: (0, -2)

 g(0) = -2 Set x  = 0.
    

Let's put it all together in a graph.

Notice the x and y intercepts are denoted by the blue dots.  The asymptotes are denoted by the broken blue lines.

  

 
Find the vertical and horizontal asymptotes, if any.  Then find x and y intercepts, if any.

[Solution]

Is there really an Asymptote?
 

Solve

            Step

     Property

Vertical Asymptote: None

  

VA: Set the denominator equal to zero. Is this really an asymptote??

Something is wrong. As x →4,  f→6.
 

This is not an asymptote but a hole in the
function at x = 4. 

After we reduce the fraction, we get
f(4) = 6,but x ≠ 4 in original function.  This means there is an hole at (4,6).

Horizontal Asymptote:  None

 as  x → ∞, f (x) = x + 2 → ∞  as  x → -∞, f (x) = x + 2 → -∞
   
Since the factor (x - 4) cancels out, we are left with the line y  = x + 2 where x ≠ 4.
 

 

 
Is there really an Asymptote?

[Solution]

 

Tutorials and Applets by
Joe McDonald
Community College of Southern Nevada