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I understand why we add or subtract from $f(x)$ to move the graph vertically. To move it up, we move all the values up by adding whatever constant value we want.

However, when it comes to horizontal shifting, we do the opposite, why is it so? Obviously I can understand how to do so (by plotting the values or using a graphing calculator, simple!). But what is the logic behind it. What if I forget how to use the other two methods and have to literally shift instead of plotting individual values of $(x, f(x))$?

Hoot
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MathEnthusiast
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  • I think of "adding something positive to $x$" as giving the graph a head start. So what used to happen further to the right on the $x$-axis is now happening earlier, at a smaller $x$-value. – pjs36 May 20 '16 at 06:00
  • I like Christian Blatter's answer here. That question has lots of other answers that you may find helpful. – Will Orrick May 20 '16 at 06:02
  • But wouldn't it be the same thing? Like if I'm plotting, say, $f(2+3)$. That equals $f(5)$ which will give us the value of the function at $x=5$ same as the original function. – MathEnthusiast May 20 '16 at 06:03
  • Outside the f (x) we are affecting the output after we have calculated so we adjusting the result. Inside the f (x) we are affecting the input before we do the input. So if we are starting at x+5 we have "push" it back by 5 to compare it with the original. – fleablood May 20 '16 at 06:17
  • I was thinking about this problem and I think I got it. Consider $f(x) = mx + b$. Now, when we say that we wanna plot $f(x+c)$, what we actually wanna plot are points $(x, f(x + c))$. Now, consider $f(x)=x$. In this case, if we want to plot at $x=2$, the value of $f(x)$ will be 2. But if we wanna plot $f(2+2)$, then the value of the y-coordinate will be 4, but x will still remain 2. In order to accomplish this, we'd have to shift the graph to the left and to the right if we're using $-2$, is my reasoning okay? – MathEnthusiast May 20 '16 at 06:23

3 Answers3

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Consider $g(x) := f(x-c)$. One can easily verify that if the point $(a,b)$ is on the graph of $f$, then the point $(a+c, b)$ is on the graph of $g$, since $g(a+c) = f((a+c) - c) = f(a) = b$.

The "deeper" reason this is true is because if one has $h(x) := (f \circ g )(x)$ where $g$ is bijective, and, furthermore, $f$ maps $a$ to $b$, then the corresponding point on the graph of $h$ will be $(g^{-1}(a), b)$. The reason is because we need to choose the input such that $g$ maps the input to $a$. Then, since it is a composite, $f$ will map $a$ to $b$, and so our final output is $b$.

MathematicsStudent1122
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  • I was thinking about this problem and I think I got it. Consider $f(x) = mx + b$. Now, when we say that we wanna plot $f(x+c)$, what we actually wanna plot are points $(x, f(x + c))$. Now, consider $f(x)=x$. In this case, if we want to plot at $x=2$, the value of $f(x)$ will be 2. But if we wanna plot $f(2+2)$, then the value of the y-coordinate will be 4, but x will still remain 2. In order to accomplish this, we'd have to shift the graph to the left and to the right if we're using $-2$, is my reasoning okay? – MathEnthusiast May 20 '16 at 06:22
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One way to think about it is like roots of an equation. Take for example the factored quadratic $(x-3)(x-2)=0$. We know that the two roots are $x=3$ and $x=2$ if we isolate each linear term and solve each for $x$. The similarity with roots and horizontal shifts is the negative value. A positive root will be expressed as a negative in its associated linear factor, and it is the same concept for horizontal shifts. If we were to solve for the $x$-intercepts of some (shifted) function, we would be isolating linear factors and solving in the same way, thus the negative is required to shift along the positive $x$-axis.

Granted, this isn't a technical explanation, but one that might make some sense intuitively.

Ethan Hunt
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Definition 0. Given $f \subseteq \mathbb{R} \times \mathbb{R}$, define $(x_0,y_0)+f \subseteq \mathbb{R} \times \mathbb{R}$ as follows:

$$(x,y) \in (x_0,y_0)+f \iff (x-x_0,y-y_0) \in f$$

(For each $(x_0,y_0) \in \mathbb{R} \times \mathbb{R}.$)

Geometrically, we can think of $(x_0,y_0)+f$ as the result of translating $f$ in the direction of the vector $(x_0,y_0)$.

Proposition 0. Suppose $f \subseteq \mathbb{R} \times \mathbb{R}$ is a function.

Then $(x_0,y_0)+f$ is also a function, and $$((x_0,y_0)+f)(x) = y_0+f(x-x_0)$$

(For each $(x_0,y_0) \in \mathbb{R} \times \mathbb{R}.$)

I can offer a proof, if you want, but I'm a little pressed for time right now.

goblin GONE
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  • I was thinking about this problem and I think I got it. Consider $f(x) = mx + b$. Now, when we say that we wanna plot $f(x+c)$, what we actually wanna plot are points $(x, f(x + c))$. Now, consider $f(x)=x$. In this case, if we want to plot at $x=2$, the value of $f(x)$ will be 2. But if we wanna plot $f(2+2)$, then the value of the y-coordinate will be 4, but x will still remain 2. In order to accomplish this, we'd have to shift the graph to the left and to the right if we're using $-2$, is my reasoning okay? – MathEnthusiast May 20 '16 at 06:23
  • @user331377, well, that's one level of analysis. Read my answer carefully for a deeper level of analysis. I'm happy to answer questions about it. – goblin GONE May 20 '16 at 06:25