Question 472207: Give the equation of the oblique asymptote, if any, of the function.
A. 
B. 
C. 
D. no oblique asymptote
Found 2 solutions by ccs2011, lwsshak3:
Answer by ccs2011(207)   (Show Source):
You can put this solution on YOUR website!
An oblique asymptote exists only when the degree of the numerator is higher than the denominator.
To find the oblique asymptote, use polynomial division and ignore the remainder.
x+3 | x^2 - 3x + 9
Step 1: (x+3) goes into x^2, x times
Step 2: Multiply x* x+3, then subtract
_____ x___________
x+3 | x^2 - 3x + 9
___ -(x^2 +3x)
-------------------
_________ -6x + 9
Step 1: (x+3) goes into -6x, -6 times
Step 2: Multiply -6* x+3, then subtract
_____ x - 6___________
x+3 | x^2 - 3x + 9
___ -(x^2 +3x)
-------------------
_________ -6x + 9
________-(-6x -18)
-------------------
______________ 27
(x+3) goes into 27, 0 times, so 27 is the remainder.
Result of the division is x-6.
Therefore the oblique asymptote is y = x - 6.
Answer by lwsshak3(11628)  (Show Source):
You can put this solution on YOUR website! Give the equation of the oblique asymptote, if any, of the function.
f(x) =(x^2-3x+9)/(x+3)
A. x = y + 3
B. y = x + 12
C. y = x - 6
D. no oblique asymptote
**
When the numerator is one degree higher than the degree of the denominator, as in this case, you will get a slant or oblique asymptote which will be a straight line of the form: y=mx+b, with m=slope and b=y-intercept.
..
To find the equation of the oblique asymptote, divide numerator by denominator. The quotient part of the answer is the equation of the oblique asymptote.
..
divide by synthetic division:
-3).....1.....-3.......9...............
.................-3......18.......
..........1....-6......27..........
f(x)=(x-6)+27
Equation of oblique asymptote: y=x-6
C is the correct ans
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