y=-log(3)x+2. Find domain, x-intercept and vertical asymptote

Question

y=-log(3)x+2. Find domain, x-intercept and vertical asymptote, I also have to sketch a graph 

Answer 1

y = log₃x^-1 + 2

y = log₃(1/x) + 2

vertical asymptote: x = 0.

 

A log always produces a number > 0.

D = {x | x > 0}

 

0 = -log₃x + 2

-2 = -log₃x

2 = log₃x

3² = x

9 = x

x-intercept: (9, 0)