What are the holes, VA, domain, HA, and Range of n+8/6n? Please help!!

We show that f(x) n+8/6n = 6 x n = 0

which flips the n+8/1 = 0+8/0-6= x = 3  this is the range.

For the HA we would work left to right.

x goes to positive or negative infinity and is determined by the highest degree terms of the polynomials in the numerator and the denominator. This particular function has polynomials of degree 0 in both the numerator and the denominator

If say n+8 was n+2 then we would use the 2/-2+3 and get 1 and show the hole as the source;

hole : -2+1 as non equal sign. but not in the case of n+8/6n

-2+1 represents 1/3 symmetry.

We see for n+8/6n  with interpreted back into the zero format minus

-0+8/-0-6 we see there is symmetry and can work on the left side of graph and flip over. Where 0 = n+8 and 1=nx6

Step-by-step explanation:

There would be no way of doing the others unless the exponents had been squared ^2

If they were squared then the domain will be (-infinity -3) parenthesis

union of( -3 -2) union of +2 to negative infinity.

There is not a vertical asymptote as the numerator divides into dominator at point 8 as a decimal.

The holes are then closed.

f(x) has value 150 at x= 0

and 0 at x= 3

so d(f(x))/dt = slope = -150/3 = -50

cross multiplying, we get :

⇒ (3) × (5) = (9) × (x)

⇒ 15 = 9x