SOLUTION: How do I graph this rational function on an oblique (slant) asymptote: T(W)=w^2+2w+4÷w-1
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Question 619989: How do I graph this rational function on an oblique (slant) asymptote: T(W)=w^2+2w+4÷w-1
Answer by lwsshak3(11628) (Show Source): You can put this solution on YOUR website!
How do I graph this rational function on an oblique (slant) asymptote:
T(W)=w^2+2w+4÷w-1
**
Given equation has a slant asymptote because the degree of the denominator is 1 less than the degree of the denominator.
To find the equation of the slant asymptote, divide by long division or synthetic division.
The resulting quotient is the straight line equation of this asymptote.
..
By synthetic division:
...1...|...1.......2.......4.............
........|............1.......3............
........|...1.......3.......7............
T(w)=(w+3) + remainder 7
Equation of slant asymptote: w+3
T(0)=-4
note: vertical asymptote at 1 (not shown on graph)
see graph below:
y=(x^2+2x+4)/(x-1)
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