How do you find the asymptotes for (x^4-3x^3-21x^2+43x+60)/(x^4-6x^3+x^2+24x-20)x4−3x3−21x2+43x+60x4−6x3+x2+24x−20?
1 Answer
Horizontal asymptote (left and right):
Vertical asymptotes at
Hole at
Explanation:
Let's attempt to factorise the numerator and denominator first.
Numerator
Substituting
1+3-21-43+60 = 01+3−21−43+60=0
So
x^4-3x^3-21x^2+43x+60 = (x+1)(x^3-4x^2-17x+60)x4−3x3−21x2+43x+60=(x+1)(x3−4x2−17x+60)
Substituting
27-36-51+60 = 027−36−51+60=0
So
x^3-4x^2-17x+60=(x-3)(x^2-x-20) = (x-3)(x-5)(x+4)x3−4x2−17x+60=(x−3)(x2−x−20)=(x−3)(x−5)(x+4)
In summary:
x^4-3x^3-21x^2+43x+60 = (x+1)(x-3)(x-5)(x+4)x4−3x3−21x2+43x+60=(x+1)(x−3)(x−5)(x+4)
Denominator
Substituting
1-6+1+24-20 = 01−6+1+24−20=0
So
x^4-6x^3+x^2+24x-20 = (x-1)(x^3-5x^2-4x+20)x4−6x3+x2+24x−20=(x−1)(x3−5x2−4x+20)
The remaining cubic factors by grouping:
x^3-5x^2-4x+20x3−5x2−4x+20
=(x^3-5x^2)-(4x-20)=(x3−5x2)−(4x−20)
=x^2(x-5)-4(x-5)=x2(x−5)−4(x−5)
=(x^2-4)(x-5)=(x2−4)(x−5)
=(x-2)(x+2)(x-5)=(x−2)(x+2)(x−5)
In summary:
x^4-6x^3+x^2+24x-20 = (x-1)(x-2)(x+2)(x-5)x4−6x3+x2+24x−20=(x−1)(x−2)(x+2)(x−5)
Together
(x^4-3x^3-21x^2+43x+60)/(x^4-6x^3+x^2+24x-20)x4−3x3−21x2+43x+60x4−6x3+x2+24x−20
=((x+1)(x-3)color(red)(cancel(color(black)((x-5))))(x+4))/((x-1)(x-2)(x+2)color(red)(cancel(color(black)((x-5)))))
=((x+1)(x-3)(x+4))/((x-1)(x-2)(x+2))
with exclusion
Looking at the leading terms of the numerator and denominator, the degrees and coefficients are identical, so there is a horizontal asymptote (left and right):
When
Substituting the value
graph{(x^4-3x^3-21x^2+43x+60)/(x^4-6x^3+x^2+24x-20) [-41.83, 38.17, -10.24, 29.76]}