Question

How do you graph `y = -1 + log_2 x`?

Answer 1

Inversely, `x=2^(y+1)>=0`. The y-axis downwards is the vertical asymptote to the graph. The curve cuts x-axis at (2, 0). As `y to oo, x to oo`. Points; `(2^(N+1), N), N=0,+-1,+-2,+-3..`

Explanation:

Use `b^y=x` is the inverse of `y=log_b x`

Here, `(y+1)=log_2 x`. So, `x=2^(y+1)`.

Give integer values for y and find plotting points `(2^(N+1), N), N=0,+-1,+-2,+-3..`

The sloping-down graph rises in the 4th quadrant, from the proximity of the asymptotic y-axis, passes through (2, 0) and in the first quadrant rises to fall flat, in the limit. As `y to oo, x to oo`. .