Question 607041
I need to find the oblique asymptote for (x^3-9x)/(x^2+x-20)
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When degree of numerator is one higher than that of the denominator, you will have an oblique asymptote.
To find the oblique asymptote, divide numerator by denominator.(by long division)
........................x..... -1.............................
....x^2+x-20 )  x^3 +0.... +9x .......................
........................x^3 +x^2-20x..................................
...............................-x^2-11x..........................
...............................-x^2    -x+20..........................
.....................................  -10x-20..................
You will get an answer=(x-1)+remainder(10x-20)
Equation of oblique asymptote: y=x-1
see graph below:
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{{{ graph( 300, 300, -20,20, -20, 20, (x^3-9x)/(x^2+x-20),x-1) }}}