Answer:
Vertical asymptotes: x=-9 and x=3
Slant asymptote: y=5x-1
Step-by-step explanation:
Given
f(x)=(5x^3+29x^2-140x+21)/(x^2+6x-27)
For vertical asymptote, the denominator is put equal to zero,so
x^2+6x-27=0
Factorizing
x^2+9x-3x-27=0
x(x+9)-3(x+9)=0
(x+9)(x-3)=0
So,
x=-9 ;x=3
As the degree of the numerator is greater than the denominator the function will not have horizontal asymptote but it will have a slant asymptote which will be calculated by long division.
After dividing 5x^3+29x^2-140x+21 by x^2+6x-27 we get
Quotient: 5x-1
Remainder: x-6
We only need the quotient for the slant asymptote,
So the slant asymptote is y = 5x -1 ..
