Find an equation of a rational function, that satisfies the given requirements.

Question

x- intercepts: (-6,0) and (2,0)

y- intercepts: (0,6)

asymptote slant about: (1/2)x+3/2

horizontal asymptote : y=-1 or vertical asymptote: x=-1

domain: (-inf,-1)u(-1,inf)

Answer 1

A rational function cannot have both a slant and horizontal asymptote so since this function has a slant asymptote we can eliminate the horizontal asymptote from consideration

The x intercepts tell us where the function equals zero so this gives us factors of (x+6) and (x-2) in the numerator

The vertical asymptote tells us where the function is undefined which is where the denominator equals zero so we have a factor of ( x+1) in the denominator

A slant asymptote occurs where the power of the numerator is greater than the power of the denominator and when we dive the numerator by the denominator we have a remainder which in this case is ( x/2 +3/2) which we will put as a factor in the numerator so we have

y = a(x+6)(x-2)(x/2+3/2)/x+1

Will use the y intercept values to solve for a

6= a(6+0)(0-2)(0/2+3/2)

6= a(-18)

a= (-1/3) so final equation is

y= (-1/3)((x+6)(x-2)(x/2+3/2)/(x-1)