Part 1
f(x) = (x-1)/(x+2)
y = (x-1)/(x+2)
x = (y-1)/(y+2)
x(y+2) = y-1
xy+2x = y-1
2x+1 = y-xy
2x+1 = y(x-1)
(2x+1)/(x-1) = y
f-1(x) = (2x+1)/(x-1)
f-1(0) = (2(0)+1)/(0-1) = 1/-1 = -1
Part 2
f(x) = (x-3)/(x+2) has a vertical asymptote at x=-2 and a horizontal asymptote at y=1 (as the numerator and denominator's degrees are the same as well as their respective coefficients). Thus, the function's domain is (-∞,-2)∪(-2,∞) and the range is (-∞,1)∪(1,∞).
Hope this helped!

AJ L.
04/04/23
Ani G.
Yeah thanks04/03/23