I My attempt to understand horizontal transformations of functions

[logicgate]

TL;DR Summary

I just want to know if my attempt is valid and correct. If it's not, then how can I improve it to make it better.

So assuming I have a graph of a parent function f(x) and I want to graph for example the function f(2x+1). I need to find a way to manipulate the function f(x) to make it look like the function f(2x+1).
For the parent function f(x) I have coordinates of (x , y).
And for the function f(2x+1) I have coordinates of (???, y).
Since both functions have the y coordinate the same, I can equate both functions : f(2x+1) = f(x).
My goal is to manipulate both sides so that the new function f(2x+1) becomes the same as the parent function.
I begin with subtracting one from both sides : f(2x) = f(x-1)
Then divide both sides by 2 : f(x) = f((x-1)/2)
This tells me that the x-coordinate of f(2x + 1) is (x-1)/2
Which means that if I want to move from parent function to f(2x+1), from every point on parent function I have to move one step to the left then divide by 2 to get into the function f(2x+1)
Is my approach valid ?

 

[PeroK]

You should recognise that $f(2x+1)$ is a new function, which is a composition of $f$ and the function $2x + 1$. I would call this function $g(x) = f(2x+1)$.

Then $g(0) = f(1)$ and we see that $g$ is $f$ shifted to the left.. Also, $g(1) = f(3)$, so $g$ is a compressed version of $f$ (horizontally compressed by a factor of 2).

That seems a conceptually clearer way to look at it. Using $f$ for both functions leads to a conceptual tangle.

 

[Mark44]

So assuming I have a graph of a parent function f(x) and I want to graph for example the function f(2x+1).

f(2x + 1) = f(2(x + 1/2)). Here we have two transformations of the graph of y = f(x): a compression toward the y-axis by a factor of 2, and a translation (a rigid shift) to the left by 1/2 unit. These transformations must be "built into" the graph of y = f(x) in that order.

Using a specific example of $f(x) = x^2$, y = f(2x) results in a thinner (i.e., compressed toward the y-axis version of the graph of y = f(x).
The graph of y = f(2(x + 1/2)) shifts all the points of the graph of y = f(2x) to the left by 1/2 unit. This can be easily verified.

Checking with a few points -- (0, 0), (1, 1), and (2, 4) lie on the graph of $y = f(x) = x^2$.
From the work above, the graph of $y = f(2x) = 4x^2$ should contain (0, 0), (1, 4), and (2, 16).

Similarly, the graph of $y = f(2x + 1) = f(2(x + 1/2))$ should contain (-1/2, 0), (1/2, 4), and (3/2, 16). One can verify that these are points on the latter graph by noting that $f(2x + 1) = 4x^2 + 4x + 1$ and substituting the three x-values of the sample points into the $4x^2 + 4x + 1$ expression to find the corresponding y-values.

 

[Mark44]

I begin with subtracting one from both sides : f(2x) = f(x-1)
Then divide both sides by 2 : f(x) = f((x-1)/2)

Neither of these is correct. First, you've started with an equation that isn't true; namely, f(x) = f(2x + 1). As pointed out by @PeroK, this is the wrong way of looking at things. Second, adding -1 to both sides doesn't result in the equation you showed. $f(2x+ 1) - 1 \ne f(2x)$. The algebra simply doesn't work that way. For example, is $\sqrt{2*4 + 1} - 1 = \sqrt 8$?

Similarly, your second equation is also wrong. $\frac 1 2 f(2x) \ne f(x)$. For example, is \frac 1 2 (2 * 2)^2 = 2^2##?

 

[logicgate]

Thanks for the answer. I have a question : Do inverse functions relate to horizontal transformations of graphs ?
Like for example the function |x-1| is treated as |x| shifted one unit to the right. Is it because the inverse of x-1 is x + 1 ?

 

[PeroK]

Thanks for the answer. I have a question : Do inverse functions relate to horizontal transformations of graphs ?
Like for example the function |x-1| is treated as |x| shifted one unit to the right. Is it because the inverse of x-1 is x + 1 ?

Inverse functions graphically are related by swapping the x and y axes.

 

[Mark44]

Do inverse functions relate to horizontal transformations of graphs ?

In general, no. Assuming that y = f(x) has an inverse, the graph of $y = f^{-1}(x)$ is the reflection across the line y = x of the graph of y = f(x). Your examples of y = x - 1 and y = x + 1 are reflections of each other across the line y = x.