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

Jun 4, 2016

Inversely, $x = {2}^{y + 1} \ge 0$. The y-axis downwards is the vertical asymptote to the graph. The curve cuts x-axis at (2, 0). As $y \to \infty , x \to \infty$. Points; $\left({2}^{N + 1} , N\right) , N = 0 , \pm 1 , \pm 2 , \pm 3. .$

Explanation:

Use ${b}^{y} = x$ is the inverse of $y = {\log}_{b} x$

Here, $\left(y + 1\right) = {\log}_{2} x$. So, $x = {2}^{y + 1}$.

Give integer values for y and find plotting points $\left({2}^{N + 1} , N\right) , N = 0 , \pm 1 , \pm 2 , \pm 3. .$

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 $y \to \infty , x \to \infty$. .