Darryl K. answered 06/14/16
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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.