4, 5] state the horizontal asymptote of the rational function. f(x) = quantity x plus nine divided by quantity x squared plus two x plus three.

the denominator polynomial has a higher degree than the numerator polynomial, so it will get large faster for large values of x. the result will tend to zero. hence the appropriate choice is

y = 0

the numerator changes sign at x=-9, but the denominator is always positive. hence the quotient will be negative for x < -9 and will be positive for x > -9. that is, the asymptote will be approached from below on the left, but from above on the right.

1. when "a" is a root of a polynomial, (x -a) is a factor of it. for the three roots given, the factors of the desired polynomial are (x +2)(x -1/2)(x -3).

in order to make the leading coefficient be negative, we need to multiply this product by a negative number. any negative number will do, but we choose a small (magnitude) value that will eliminate the fraction: -2.

then

f(x) = -2(x +2)(x -1/2)(x -3) = -(x +2)(2x -1)(x -3)

= -(2x² +3x -2)(x -3)

= -(2x³ -3x² -11x +6)

f(x) = -2x³ +3x² +11x -6

2. a graph created by the desmos on-line graphing calculator is shown, and the zeros are highlighted.

3. as indicated in part 1, the multiplier of this equation can be anything and the zeros will remain the same. you want a negative leading coefficient, so the "anything" is restricted to any of the infinite number of numbers that will make that be the case.