Why does adding a constant to the x value in a graph move it to the left?

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Adding a positive constant to the x value in a function shifts the graph to the left because it effectively changes the input required to reach the same output. For example, in the function f(x) = x^2, modifying it to g(x) = (x+c)^2 means the zero of the function moves from x=0 to x=-c. This shift occurs because the modified function requires a smaller x value to achieve the same output as the original function. Consequently, for any value b, the modified function g(x) will reach that value at x=b-c, illustrating the leftward movement. Understanding this concept clarifies how horizontal shifts in graphs operate in relation to constant additions.
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In graphing functions, I don't get how adding a constant c to the x value in say y=f(x) moves the graph to the left. Surely, if you add a constant to x, its value increase and so the graph should move to the right?
 
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The graph moves to the left if you add a positive constant.
To see why this is so I personally like to think
of the following example:Lets take the function f(x) = x^2.

Where is the function zero?
It is zero at x=0.

Now let's take the modified function g(x) = f(x+c) = (x+c)^2
(with c being positive).
Where is this function zero?
The function is zero at x=-c.

Therefore, the zero of the function has moved to the left.
This is not only the case for the zero of the function.
You can take any arbitrary value, for example f(b)=b^2:
The function f(x) has the value b^2 at x=b.
The modified function g(x) has the value b^2 at x=b-c.Also, draw some functions and the corresponding modified functions.
For example:
f(x)=x^2 and g(x)=(x+1)^2.
 
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Lets say we have a new function f(x+c) that wants to attain the same value of f(x) for a predefined value. The new function has to work c units less to attain that value. Or, for a negative constant, it has to work c units harder, so it takes c units longer along to x-axis to reach the same values.