Building an equation with asymptotes, intercepts, and points

Question

Find an equation of a rational function f that satisfies the given conditions.
vertical asymptotes: x = −1, x = 0
horizontal asymptote: y = 0
x-intercept: 1; f(2) = 1

Answer 1

Vertical asymptotes are found by setting denomintor factors equal to zero.

 

x(x + 1)

 

 

Since the horizontal asymptote is y=0, the degree of numerator is less than degree of denominator.

 

x-intercept is the value that makes the numerator zero.  Its factor is (x - 1).

 

 

The initial value condition allows you to find the leading coefficient.

 

So here is our function so far.

 

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

 

1 = C(2 - 1) / [2(2 + 1)]

 

1 = C / 6

 

C = 6

 

Therefore, your function is

 

f(x) = 6x / (x² + x)