A function f has a horizontal asymptote of y = -4, a vertical asymptote of x = 3, and an x-intercept at (1,0).

A function f has a horizontal asymptote of y = -4, a vertical asymptote of x = 3, and an x-intercept at (1,0). PART A: Let f be of the form

                       f(x)=(ax+b)/(x+c)

                       Find an expression for f(x).

Since we have a vertical asymptote at x = 3, c  = - 3.....and since we have a horizontal asymptote at y  = -4, a  = -4......and since we have an  x intercept at (1,0), we can solve this for b ⇒  -4(1)  +  b =  0 ⇒  b  = 4......here's the graph : https://www.desmos.com/calculator/1pmoaiuapl

PART B: Let f be of the form

                       f(x)=(rx+s)/(2x+t)

Find an expression for f(x).

Since the vertical asymptote is at 3 , we can solve this for t.....2(3) + t  = 0 ⇒  t  = -6

And since the horizontal asymptote is at -4, r  = -8

And since we have an x intercept at (1,0), we can solve this for s......-8(1) + s  = 0 ⇒  s  = 8

Here's a graph : https://www.desmos.com/calculator/gyqo2l6jdl

Thanks!