A graphing question in (pre)-calculus I saw this question in a calculus textbook: "Graph $ln(1-x/2)$ using the transformation rules.
Here's my attempt:
1) We know how $\ln(x)$ looks like
2) $\ln(x/2)$ is $\ln(x)$ stretched horizontally by a factor of 2
3) $\ln(-x/2)$ is $\ln(x/2)$ reflected over the $y$-axis
4) Finally $\ln(1-x/2)$ is $\ln(-x/2)$ moved to the left 1 unit, just like $|x+1|$ is $|x|$ moved to the left 1 unit.
However, when I check with a graphing calculator, the result turned out to be wrong? I know that I was wrong in the last step, but I don't know why. I applied precisely the rules of transformation stated in the textbook. Thanks for help!
 A: $\displaystyle y=\ln\frac{-x}{2}$ moved to the left $1$ unit is $\displaystyle y=\ln\frac{-(x+1)}{2}=\ln\frac{-x-1}{2}$
A: Here is why it's not moved to the left by one unit.
Lets try to turn the equation $y = \ln(1-\frac{x}{2})$ inside-out.
$$y = \ln (1-\frac{x}{2})$$
$$\exp(y) = \exp(\ln (1-\frac{x}{2}))$$
$$\exp(y) = 1-\frac{x}{2}$$
$$2\exp(y) = 2-x$$
$$-2\exp(y) = x-2$$
$$-2\exp(y) + 2 = x$$
Or
$$x = -2e^{y} + 2$$
This graph is stretched by a factor of 2 , flipped, and shifted positively by two units in the direction of the x-axis (notice that I avoided using the words "vertical" and "horizontal").
Turning the equation back right-side-out, we have
$$-2\exp(y) +2 = x$$
$$-2\exp(y) = x-2$$
$$\exp(y) = -\frac{1}{2}(x-2)$$
$$y = \ln(-\frac{1}{2}(x-2))$$
Graphing this function, we find the same thing .... that ....
This graph is stretched by a factor of 2 , flipped, and shifted positively by two units in the direction of the x-axis (notice that I avoided using the words "vertical" and "horizontal").
This also explains why it is preferable to express the argument of a function as
$$\frac{1}{a}(x-c)$$
