Identify the asymptotes, domain, and range of the function. f(x)= 3/ (x + 2) + 1

Question

f(x)=3/(x+2)+1 

Answer 1

Step 1) Combine the two terms.

 

f(x) = 3/(x+2) + 1(x+2)/(x+2) = (3+x+2)/(x+2) = (x+5)/(x+2)

 

Step 2) Find the vertical asymptote.  Set the denominator equal to zero and solve for x.

 

x+2=0 → x = -2

 

Step 3) Find the horizontal asymptote.

 

a) If the numerator degree is smaller than the denominator degree then the horizontal asymptote is y = 0.

 

b) If the numerator and denominator degree is the same f(X) = (ax^n + …)/(bx^n+...) then the horizontal asymptote is y=a/c.

 

y=1/1 → y=1

 

Step 3) Find the domain.  The domain is all the allowable values that x can take on.  We look for two restrictions.  1) Cannot divide by zero and 2) Cannot have a negative under an even root radical.

 

Set the denominator not equal to zero and solve for x.

 

x+2≠0 → x ≠ -2

 

This is the only value x cannot be so our domain is all reals except -2.  In set builder notation we have D = {x|x≠0}.  In interval notation x = (-∞, 2) U (2, ∞).  Since the problem did not state a format for the listing the domain I listed several ways to express the domain.

 

Step 4) The range of a function is usually determined by graphing the equation which I cannot do on this platform.  Most students use a graphing calculator.  The range is all allowable values that y can be.  You can make a graph by hand by making an xy-table.  Pick points close to 2 on the left and right sides and points far away form 2.

 

x     -10     -8     -6     -4     -2.5    -2.1     -2.001     -1.999     -1.9    -1.5        0     2       4        6          8     10

 

y    0.625   0.5    0.25   -0.5  -5    -29         -2999      3001       31      7       2.5   1.75    1.5    1.375    1.3     1.25

 

From the list we see that the closer we get to -2 form the left the more negative our y-value becomes.  As we approach -2 form the left the y-value goes to negative infinity.  As we approach -2 from the right our y-value goes to positive infinity.  As x goes to infinity our y value approaches one from the top and as x goes to negative infinity our y value approached 1 form the bottom.  So it appears that our range is all y-values except 1.  Not a graph can cross a horizontal asymptote.  How do you know for sure that the graph does not cross a horizontal asymptoe?  Just set y = 1 in the original equation and solve for x.  If you can solve for x then the graph crosses its horizontal asymptote otherwise it does not.  Try this with your equation.