Understanding the graphical effect of f(x+a)

[Aion96]

Homework Statement

I'm trying to understand the transformation of a function when replacing x with x+a.

Relevant Equations

None

Solution attempt. I know that f(x+a) shifts the function f(x) by a units to the left for a>1, however, I find this counterintuitive because adding a to x feels like it should move the graph to the right rather than to the left. I understand the algebra, but I struggle with the intuition behind it.

Given a function f(x), we know that a point on its graph is (x,f(x)). Now, if we consider f(x+a), I want to determine the new corresponding points. My reasoning is:

1. The point (x+a,f(x+a) lies on the graph of the original function f(x).
2. The goal is to find the point (x,f(x+a)), and we notice that its obtained by moving the point (x+a,f(x+a) a units the left.
3. To express the transformation using x instead of x+a, we rewrite the point as (x,f(x+a)).

This suggests that the graph of f(x+a) is obtained by shifting the graph of f(x) a units to the left. I believe my reasoning is correct, but I want to confirm:

• Is my step-by-step logic sound?
• Could someone provide an intuitive explanation or a different approach to understanding this shift?

 

[fresh_42]

Yes, you are right. The problem you are struggling with, and I often, too, is to distinguish between the movement of the object (the function graph) and the movement of the coordinate system, here the $x$-axis. Moving the function to the left is the same as shifting the coordinate system to the right. Other than doing it on a table, we do not have an exterior coordinate system (the table) here to distinguish between both.

 

[FactChecker]

Homework Statement: I'm trying to understand the transformation of a function when replacing x with x+a.
Relevant Equations: None

Solution attempt. I know that f(x+a) shifts the function f(x) by a units to the left for a>1, however, I find this counterintuitive because adding a to x feels like it should move the graph to the right rather than to the left. I understand the algebra, but I struggle with the intuition behind it.

Good question. Adding $a$ moves the value $x+a$ to the right, but not $x$ itself. The graph is based on $x$, which can start farther to the left. Then its value is shifted to the right before evaluating $f(x+a)$.

 

[Mark44]

I know that f(x+a) shifts the function f(x) by a units to the left for a>1

That would be for a > 0, not a > 1.
A more correct restatement of what you wrote is this: the graph of y = f(x + a), with a > 0, is shifted to the left by a units, relative to the graph of y = f(x).

 

[Steve4Physics]

Solution attempt. I know that f(x+a) shifts the function f(x) by a units to the left for a>1

You mean for $a \ge 0$. There's nothing wrong with, for example, $f(x + \frac 13)$ which shifts the graph $\frac 13$ of a unit left. [Edit: aha, @Mark44 beat me to it!]

One thing I (a non-mathematician) found helpful when first meeting this - a rather long time ago -was to consider $y = x$ and $y=x+1$. Sketching the two graphs immediately shows that replacing $x$ by $x+1$ shifts the first graph 1 unit left.

By considering the point where each of these graphs crosses the x-axis, it isn't too difficult see why this happens,

Of course $y=x-1$ produces a right shift by 1 unit.

I found it useful to memorise the directions of the shifts for $f(x \pm a)$. This saves time in exams and reduces the risk of a mistake.

(FWIW, the first time I found this really useful was understanding the basic wave equation $y(x,t) = A\sin(kx - \omega t)$. The shape is a simple sine curve $y = A\sin(kx)$ and this shape steadily moves right as the $\omega t$ term steadily increases, giving a moving sine curve. Apologies for sneaking some physics into a maths thread!)

 

[Mark44]

You mean for $a \ge 0$.

If a = 0, there is no shift at all. That's why I didn't include 0.

FWIW, the first time I found this really useful was understanding the basic wave equation $y(x,t) = A\sin(kx - \omega t)$. The shape is a simple sine curve $y = A\sin(kx)$ and this shape steadily moves right as the $\omega t$ term steadily increases, giving a moving sine curve. Apologies for sneaking some physics into a maths thread!)

The equation might have to do with wave equations in physics, but the underlying concepts are from mathematics.
The equation can be rewritten as $y(x,t) = A\sin(k(x - \frac \omega k t))$. This represents two transformations: a compression of the graph toward the y axis by a factor of k (if k > 1), followed by a translation to the left by $\frac \omega k$ (if that fraction is positive). If 0 < k < 1, there is an expansion away from the y axis. Note that these transformations must be performed in the order I showed, or the resulting graph won't be correct.

 

[FactChecker]

If you interpret a shift to the left by a negative amount to be a shift to the right by a positive amount, you don't need to worry about the sign of $a$. :-)

 

[Steve4Physics]

If a = 0, there is no shift at all. That's why I didn't include 0.

Yes. I thought about that at the time. On reflection (no pun intended) I wish I had used $a \gt 0$. I hadn't seen your post at the time.

The equation might have to do with wave equations in physics, but the underlying concepts are from mathematics.

Of course! I only pointed out the wave equation example because it brought back happy memories of something I realised/discovered for myself. And, in the post, I did explicitly apologise for sneaking-in some physics!

 

[Gavran]

• Could someone provide an intuitive explanation or a different approach to understanding this shift?

When you shift the graph $G_1=\{(x,f(x))|x\in X\}$ to the left by $a$ units you will get the graph $G_2=\{(x-a,f(x))|x\in X\}$. The graph $G_2=\{(x-a,f(x))|x\in X\}$ is exactly the same as the graph $\{(x,f(x+a))|(x+a)\in X\}$.

 

[FactChecker]

The graph $G_2=\{(x-a,f(x))|x\in X\}$ is exactly the same as the graph $\{(x,f(x+a))|(x+a)\in X\}$.

The sets are identical, but the graphs are not. As sets, $x$ is a dummy variable that disappears. As a graph of $x$ versus $f(x)$, $x=0$ sets the origin, and the origins of those two are not the same.

 

[Bling Fizikst]

Not a rigorous post . But how i interpret $g(x)=f(x+a)$ is just setting the argument of $f$ to zero ,i.e, $x+a=0$ this means : $x=-a$ . So , whatever graph you have for $f(x)$ , you get the same value of $f(0)$ at $x=-a$,i.e, $g(-a)=f(-a+a)=f(0)$ ($a>0$) . So , it is as good as shifting the graph of $f$ left by $a$ units . Similarly , you can try other values of $f$ to get a 'feel' . But i have just made a trick like this so that i don't waste time .

Let's say : $f(x)=x$ , then $g(x)=f(x+3)=x+3$ , So, $g(0-3)=f(0)$

Your $f(0)$ value is shifted by $3$ units to the left for $g$ , so are all the other values . Hence , you can say $g(x)$ is the graph of $f(x)$ shifted by $3$ units to the left

 

[neilparker62]

Some will say it's arbitrary but I think the correct formula for a horizontal shift is $f(x-a)$ and not $f(x+a)$ in the first place. The former lines up with the op's intuition (mine too!) which is that a positive value of a will shift the graph to the right. I have old text books which write (for example) the equation of parabolas as $f(x) = a(x-p)^2+q$ whereas our syllabus document uses $f(x)=a(x+p)^2+q$.

 

[Mark44]

Some will say it's arbitrary but I think the correct formula for a horizontal shift is $f(x-a)$ and not $f(x+a)$ in the first place.

It all depends on whether we constrain a to be a positive real number.
In your parabola example, if p > 0 and q > 0, the graph of $f(x) = a(x + p)^2 + q$ will be translations to the left by p and up by p q of the graph of $y = ax^2$

 

[neilparker62]

It all depends on whether we constrain a to be a positive real number.
In your parabola example, if p > 0 and q > 0, the graph of $f(x) = a(x + p)^2 + q$ will be translations to the left by p and up by p of the graph of $y = ax^2$

Yes - it's intended that a is a real number but not necessarily positive. For example f(x-(-2)) would represent a left shift of 2 units on the graph of f(x).

In the parabola example, the problem with p>0 creating a left shift is that it is counter-intuitive as per the op's original observation. A positive value of p should result in a right shift since that is the positive direction on the horizontal x-axis. So I'd go with the old text books:

Edit: Replaced image from text book with one which includes a useful explanation.

 

[Mark44]

In the parabola example, the problem with p>0 creating a left shift is that it is counter-intuitive as per the op's original observation. A positive value of p should result in a right shift since that is the positive direction on the horizontal x-axis.

Before people learn better, their intuition can be wrong... A purpose of education is to correct erroneous intuitions.

 

[FactChecker]

There is no police force that will enforce an official "Horizontal Graph Shifting Rule". If a student is initially confused about right/left shift for a positive/negative $a$, he should be able to figure it out and accept the result. There are more difficult concepts ahead.

 

[neilparker62]

Before people learn better, their intuition can be wrong... A purpose of education is to correct erroneous intuitions.

If we go by the horizontal shifting rule which I have imaged above, the op's intuition is completely correct. A positive value of h leads to a right shift and a negative value of h leads to a left shift. On the other hand if we prefer a convention that writes the shifting rule as F(x) = f(x + h), then an exact "counter-intuition" applies whereby positive values of h lead to left shifts and and negative values of h lead to right shifts.

Similarly a student could be puzzled because his/her reasonable intuition indicates that electrons flow from negative to positive whereas there is an established convention that current flows from positive to negative. There is - however - nothing particularly wrong with the student's intuition in this instance.

Agreed that intuition can be wrong - as is evidenced by Aristotle's erroneous belief that heavier objects fall faster than light ones. Corrected by Galileo.

 

[Gavran]

left by p and up by p

left by p and up by q

 

[Mark44]

left by p and up by q

Right, and that's what I meant. Fixed in the original

 

[Mark44]

Similarly a student could be puzzled because his/her reasonable intuition indicates that electrons flow from negative to positive whereas there is an established convention that current flows from positive to negative.

I'm not sure this is a comparable example. The labels "positive" and "negative" came about in the first place, I believe, because of a faulty understanding in how electrons flowed in a circuit.