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 [tex] y=\frac{7}{2}[/tex].
5.No, horizontal asymptote.
6.Oblique asymptote at [tex] y= x-6 [/tex].
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, [tex]x^2+3x-10=0[/tex]
Factorize the polynomial we get
[tex](x+5)(x-2)=0[/tex]
[tex] x+5=0 [/tex] and [tex] x-2=0[/tex]
[tex] x=-5[/tex] and [tex] x=2[/tex]
Hence, the vertical asymptote of f(x) at x= -5 and x=2
2.[tex] x^2-9x+18=0[/tex]
Fatorize the polynomial then we get
[tex] (x-6)(x-3)=0[/tex]
[tex]x-6=0[/tex] and [tex]x-3=0[/tex]
[tex] x=6 [/tex] and [tex]x=3[/tex].
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=[tex]\frac{7}{2}[/tex].
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
