Am I understanding horizontal shift properly?

[r0bHadz]

Homework Statement

I mean, it does make sense, if I plot f(x)= x^2 and f(x-1) = (x-1)^2 and find where they intersect the x/y axis, it does make sense. But the book I'm reading (Stewarts) confuses me with the following:"Likewise, if g(x) = f(x-c), c>0, then the value of g at x is the same as the value of f at x-c. Therefore, the graph of y= f(x-c) is just the graph of y=f(x) shifted c units to the right"

Homework Equations

The Attempt at a Solution

The first part does not confuse me at all. If you have g(x), and at the point x it is the same value as the function f at the point x-c, then it will be the same function f, shifted to the right c units.

For the second part please let me know if I am understanding correctly:

It would be like starting with just your graph of the coordinate system, y and x, and then compressing x like it is a spring, by c units. then you draw your function, and let go of he spring. now you have your original axis, with the function shifted and the max height of the spring is where your function is suppose to be graphed on your original axis

 

[r0bHadz]

Hmm it seems like its best to just look at it in terms of a number line. x-3 = 0 means x is located at location 3 despite the -.

 

[Ray Vickson]

Homework Statement

I mean, it does make sense, if I plot f(x)= x^2 and f(x-1) = (x-1)^2 and find where they intersect the x/y axis, it does make sense. But the book I'm reading (Stewarts) confuses me with the following:"Likewise, if g(x) = f(x-c), c>0, then the value of g at x is the same as the value of f at x-c. Therefore, the graph of y= f(x-c) is just the graph of y=f(x) shifted c units to the right"

Homework Equations

The Attempt at a Solution

The first part does not confuse me at all. If you have g(x), and at the point x it is the same value as the function f at the point x-c, then it will be the same function f, shifted to the right c units.

For the second part please let me know if I am understanding correctly:

It would be like starting with just your graph of the coordinate system, y and x, and then compressing x like it is a spring, by c units. then you draw your function, and let go of he spring. now you have your original axis, with the function shifted and the max height of the spring is where your function is suppose to be graphed on your original axis

You can easily figure all this out for yourself---and doing that is the very best way to learn! Just take some simple $f(x)$ and plot the graphs of $y = f(x)$ and $y = f(x-c)$ for some numerical value of $c$ (such as $c=1$ or $c = -1$). When you do that, what do you see?

 

[MattS134]

Homework Statement

I mean, it does make sense, if I plot f(x)= x^2 and f(x-1) = (x-1)^2 and find where they intersect the x/y axis, it does make sense. But the book I'm reading (Stewarts) confuses me with the following:"Likewise, if g(x) = f(x-c), c>0, then the value of g at x is the same as the value of f at x-c. Therefore, the graph of y= f(x-c) is just the graph of y=f(x) shifted c units to the right"

Homework Equations

The Attempt at a Solution

The first part does not confuse me at all. If you have g(x), and at the point x it is the same value as the function f at the point x-c, then it will be the same function f, shifted to the right c units.

For the second part please let me know if I am understanding correctly:

It would be like starting with just your graph of the coordinate system, y and x, and then compressing x like it is a spring, by c units. then you draw your function, and let go of he spring. now you have your original axis, with the function shifted and the max height of the spring is where your function is suppose to be graphed on your original axis

I don't believe the graph is compressed like a spring as you originally stated, but rather shifted to the right without changing any other properties of the graph. Compressions come up when f(x) is multiplied by constants, not when they are added or subtracted