vertical asymptote at x=3
1/(x-3)
hole at x=1
(x-1)/((x-3)(x-1))
This has a horizontal asymptote at y=0. So, to get a
horizontal asymptote at y=2,
(x-1)/((x-3)(x-1)) + 2
The problem remaining is that this goes negative for x<3. So, square (x-3) and it stays positive
(x-1)/((x-3)^2(x-1)) + 2
Check the graph at
https://www.wolframalpha.com/input/?i=%28x-1%29%2F%28%28x-3%29%5E2%28x-1%29%29+%2B+2
20. (4 points) Write an equation for a rational function whose graph has all of the indicated features. vertical asymptote with equation x 3 horizontal asymptote with equation y =2 hole at x= 1 no x-intercepts
1 answer
1 answer