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.

okay so 15.15 *()5: +5/6

step-by-step explanation:

< coa = 150 ( given)

as ba and bc are tangents

so oc and oa is perpendicular to cb and ab respectively because in circle normal passes through centre

so in quadrilateral oabc

< coa + < oab + < ocb + < abc = 360

150 +90+90 + < abc = 360

330 + < abc = 360

< abc = 360 -330 = 30

so < abc = 30

⭐dhruv⭐