What’s the factored form for the rational functions?

Question

1) Conditions: Two factors in numerator, two factors in denominator. Two negative x-intercepts. No holes. A horizontal asymptote of y = 2. One vertical asymptote is located to the right of the origin and one vertical asymptote is located to the left of the origin

2) Conditions: Two factors in numerator, one factor in denominator. Both a positive and negative x-intercept. No holes. Vertical asymptote resides in quadrants 2 and 3.   

3) Conditions: Three factors in numerator and four factors in denominator. Two holes. A negative y-intercept.    

Answer 1

1)

"Two factors in numerator, two factors in denominator" implies

f(x) = K(x-a)(x-b)/(x-p)(x-q)

"Two negative x-intercepts" implies

a ≠ b, a < 0, b < 0

"No holes" implies

a ≠ p, a ≠ q, b ≠ p, b ≠ q

"A horizontal asymptote of y = 2" implies

K = 2

"One vertical asymptote is located to the right of the origin and one vertical asymptote is located to the left of the origin" implies

p > 0, q < 0

One possible solution is a = -1, b = -2, p = 1, q = -3

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

2)

"Two factors in numerator, one factor in denominator." implies

f(x) = K(x-a)(x-b)/(x-p)

"Both a positive and negative x-intercept." implies

a > 0, b < 0

"No holes." implies

a ≠ p, b ≠ p

"Vertical asymptote resides in quadrants 2 and 3." implies

p < 0

One possible solution is K = 1, a = 1, b = -2, p = -1

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

3)

"Three factors in numerator and four factors in denominator." implies

f(x) = K(x-a)(x-b)(x-c)/(x-p)(x-q)(x-r)(x-s)

"Two holes." implies

a = p, b = q, c ≠ r, c ≠ s

"A negative y-intercept." implies

p ≠ 0, q ≠ 0, r ≠ 0, s ≠ 0,

(0-c)/(0-r)(0-s) < 0, which implies that

the set {c, r, s} contains an odd number of negative values.

One possible solution is K = 1, a = -1, b = 1, c = -2, r = 2, s = 3

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