How do you graph y=1+log2x?

1 Answer
Jun 4, 2016

Inversely, x=2y+10. The y-axis downwards is the vertical asymptote to the graph. The curve cuts x-axis at (2, 0). As y,x. Points; (2N+1,N),N=0,±1,±2,±3..

Explanation:

Use by=x is the inverse of y=logbx

Here, (y+1)=log2x. So, x=2y+1.

Give integer values for y and find plotting points (2N+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,x. .