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

1 Answer
Jun 4, 2016

Answer:

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#. .