Why would $(x+1)$ transform the graph to the left?

Question

Why do transformations for the $x$ variable in graphs work opposite as you would expect?

Example: $f(x) =(x)-1$ moves the graph down as you would expect but $g(x) = (x+1)-1$ moves the graph down (as it looks like) but instead of to the right one it's to the left one.

Note: I understand fully how to do graph transformations but don't understand why we just deem the x-value transformations as "doing the opposite". Thanks!

Answer 1

Look at the objects in front of yourself. Now take a step to the right. The objects shifted to the left, not right, right?

Answer 2

Because your domain values will decrease by $1$ in order to provide the same output. Let’s say for $f(x)$, we have the following points: $$(x_1, y_1), (x_2, y_2), (x_3, y_3)$$ Now, if we have $f(x+1)$, our inputs will all decrease by one. $$(x_1-1, y_1), (x_2-1, y_2), (x_3-1, y_3)$$ Obviously, the $-1$ balances the $+1$ in $f(x+1)$. Therefore, this explains why we carry out the opposite operation to those in the parentheses. In a more general form, we get the following rule.

For positive values of $n$, shift $n$ units left for $f(x+n)$ and shift $n$ units right for $f(x-n)$.

For multiplication and division within the parentheses, the idea is exactly the same.

For values of $n$, there will be horizontal stretch by scale factor $\vert{\frac{1}{n}}\vert$ for $f(nx)$, and there will be horizontal stretch by scale factor $\vert n\vert$ for $f(\frac{x}{n})$.

Answer 3

One thing to notice is that in an equation like

$$y=(x+1)^2 - 2$$

the 1 is with the $x$, but the $-2$ is on the opposite side of the equal sign from $y$. This $-2$ moves the graph down, but put it on the other side of the equation and you have

$$y+2 = (x+1)^2.$$

So things are more symmetric than you thought. Pluses move things in the negative direction.

Next, if you think of a graph drawing machine that's going to draw the graph of $y=f(x)$ from left to right, then what happens when you increase $x$ by $1$? Everything happens 1 unit sooner. $x$ may be at $7$, but the function is being fed $8$, so the point that used to be at $8$ is now at $7$. Everything is shifted left.