1. identify the vertical asymptotes of f(x) = 2/x^2+3x-10
2. identify the vertical asymptotes of f(x) = x+6/x^2-9x +18
3. identify the horizontal asymptote of f(x) =4x/7
4. identify the horizontal asymptote of f(x) = 7x+1/2x-9
5. identify the horizontal asymptote of f(x) = x^2+5x-3/4x-1
6. identify the oblique asymptote of f(x) = x^2-4x+8 /x+2
7.identify the oblique asymptote of f(x) = 2x^2+3x+8/x+3
8.identify the oblique asymptote of f(x) = x+4/3x^2+5x-2
9. identify the oblique asymptote of f(x) =4x^2-x+2/x+1

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,

Factorize the polynomial we get

and

and

Hence, the vertical asymptote of f(x) at x= -5 and x=2

2.

Fatorize the polynomial then we get

and

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

the equation cos(35) =a/25 can be used to find the length of bc what is the length of bc? round to the nearest tenth

step-by-step explanation: