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