Answer:
1.Vertical asymptotes are at x=-5 and x=2 .
2.Vertical asymptotes are at x=6 and x=3.
3.No , horizontal asymptotes.
4.Horizontal asymptote
.
5.No, horizontal asymptote.
6.Oblique asymptote at
.
7.Oblique asymptote at y=2x-3
8.There is no oblique asymptote.
9.Oblique asymptote y=4x-5.
Step-by-step explanation:
1. To find vertical asymptote
Substitute denominator is equal to zero
Therefore, ![x^2+3x-10=0](https://tex.z-dn.net/?f=x%5E2%2B3x-10%3D0)
Factorize the polynomial we get
![(x+5)(x-2)=0](https://tex.z-dn.net/?f=%28x%2B5%29%28x-2%29%3D0)
and ![x-2=0](https://tex.z-dn.net/?f=%20x-2%3D0)
and ![x=2](https://tex.z-dn.net/?f=%20x%3D2)
Hence, the vertical asymptote of f(x) at x= -5 and x=2
2.![x^2-9x+18=0](https://tex.z-dn.net/?f=%20x%5E2-9x%2B18%3D0)
Fatorize the polynomial then we get
![(x-6)(x-3)=0](https://tex.z-dn.net/?f=%20%28x-6%29%28x-3%29%3D0)
and ![x-3=0](https://tex.z-dn.net/?f=x-3%3D0)
and
.
Therefore, the vertical asymptote of f(x) at x=6 and x=3
3.There is no horizontal asymptote of f(x) because the degree of numerator is greater than the degree of denominator.
4.Horizontal asymptote : The degree of numerator and degree of denominator are same therefore,
Horizontal asymptote=
.
5.No horizontal asymptote because the degree of numerator is greater than the degree of denominator.
6.Oblique asymptote: oblique asymptote is the value of quotient which obtained by dividing the numerator by denominator .
Oblique asymptote of f(x) at y=x-6
7. Oblique asymptote of f(x) at y= 2x-3
8. There is no oblique asymptote because the degree of numerator is less than the degree of denominator.
9.Oblique asymptote of f(x) at y= 4x-5