Assume a function called $f(x)$ , Then all of us know that of we draw $f(x+a)$ it will be a transformation to the left or right and $f(x)+b$ to up or down.

But when I drew floor function on Desmos online graphing I found something a little bit different.


As you see that the black function is exactly on the purple one and that confusing me , Here $f(x+a)=f(x)+a$ , And we are just rising the function up .

To be more specific I need you to explain these question :

$1.$ Why when we add a number in the floor function notation $f(x+a)$ or $[\frac {x}{2}+1]$ it rise the function.

$2.$How can transform the function to the left just a unit.


The fact that $\lfloor f(x) + 1 \rfloor = \lfloor f(x) \rfloor + 1$ for any function $f$ should be fairly obvious, if you think about what the floor function does: round down to nearest integer. (You can replace "$+1$" here with "$+n$" for any integer $n$, but it doesn't work for non-integers.)

So for your particular example $f(x) = \frac{x}{2}$, you get $\lfloor \frac{x}{2} + 1 \rfloor = \lfloor \frac{x}{2} \rfloor + 1$, and the graph is moved one unit upwards.

But that's also equal to $\lfloor \frac{x}{2} + 1 \rfloor = \lfloor \frac{x+2}{2} \rfloor = \lfloor f(x+2) \rfloor$, so it's the same thing as moving the graph two units to the left.

If you want to move the graph one unit to the left, take $\lfloor f(x+1) \rfloor = \lfloor \frac{x+1}{2} \rfloor = \lfloor \frac{x}{2} + \frac{1}{2} \rfloor$ instead.

  • $\begingroup$ When we say $[\frac {x}{2}+1]$ is this trassfomation to left or up ? $\endgroup$ – Mohammad Alshareef May 7 at 10:51
  • $\begingroup$ @MohammadAlshareef: Both! As I wrote, it's the same thing (for this particular graph). $\endgroup$ – Hans Lundmark May 7 at 11:38

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.