horizantal shifts of graphs: why do i have to shift to the right if...

I

Illvoices

Junior Member
Joined
Jan 13, 2017
Messages
116
why do i have to shift to the right if the symbol is negative in "To graph y = f(x-c)"
 
Illvoices said:
why do i have to shift to the right if the symbol is negative in "To graph y = f(x-c)"
Click to expand...

Consider, for example the graph of \(\displaystyle y=x^2\)

This goes through the points (-1, 1),(0, 0), (1, 1), (2, 4) and (3, 9) etc. Right?

Now consider the graph of \(\displaystyle y=(x-1)^2\)

This passes through the points (0,1), (1,0), (2, 1), (3, 4) (4, 9) etc.

Now compare these points to the ones above. The points have all moved 1 unit to the right.

Look at the point (3, 9) on the graph of \(\displaystyle y=x^2\). This is because \(\displaystyle 3^2 =9\).

Now look at the point (4, 9) on the graph of \(\displaystyle y=(x-1)^2\). This is because \(\displaystyle (4-1)^2 =9\). See how x has to equal 4 here to get the same result (ie y=9) as x=3 got in the first graph. That is a movement to the right.

Now try it for yourself with \(\displaystyle y=(x+1)^2\) to convince yourself why that moves to the LEFT.
 
Illvoices said:
why do i have to shift to the right if the symbol is negative in "To graph y = f(x-c)"
Click to expand...

A short answer is: "Because that's the rule."

A long answer helps to explain why, but it requires more thought. (What follows are just two ways of considering it.)

Let's compare two simple functions, h and g:

h(x) = x - 1

g(x) = x

If you compare the graphs (or compare tables of values) for functions h and g, you can convince yourself that the graph of h shows what g is doing when x is one unit "earlier".

For examples:

When x is at 3, h is 2. Well, g is also 2, but that happens when x is 2 (one unit "earlier" than when x is 3).

When x grows to 4, h grows to 3. Function g is 3 when x is 3 (one unit before 4).

h = 4 when x = 5
g = 4 when x = 4 (one unit before 5)

h = 5 when x = 6
g = 5 when x = 5 (one unit before 6)

In other words, function h's behavior matches what function g is doing one unit earlier. Therefore, you want to shift the graph of g(x) to the right, for it to "catch up" to h(x).

Another approach to thinking about this is to realize that coordinate x-1 is one unit to the left of coordinate x.

If the entire coordinate system shifts one unit to the left, but graphs do not shift with it, then the net effect is that the graphs become shifted one unit to the right (with respect to the axes).

Imagine the graph of g drawn on a transparency, laying on top of graph paper. If you slide the graph paper to the left one unit (all x become x-1) without also sliding the transparency, then you'll "see" the graph of g shifting to the right one unit (ending up where the graph of h lives, on an unshifted coordinate system.)

It takes students some time to wrap their mind around all of this, so, if it still seems fuzzy, then you could just accept and use the rule for now (subtracting a positive constant from all inputs shifts the graph to the right, and adding a positive constant to all inputs shifts the graph to the left). Your comprehension of why it works this way will come with more experience. :cool:
 
Last edited:
Illvoices said:
why do i have to shift to the right if the symbol is negative in "To graph y = f(x-c)"
Click to expand...

Consider f(x) = x2 and h(x) = (x-3)2.

Now, for example, when you square 5 you get 25. It does not matter how you came to be squaring 5. You might have computed 4+1 and when you square it you get 25. You might have computed 7-2 and when you square that you get 25. This concept is true with other numbers as well (ie (3+4)2=(2+5)2=49 and for other functions as well-think about that one)

Now back to f(x) and h(x). f(7)= 72=49. Now to be squaring 7 in the h(x) function you need to let x be 3 more because you are subtracting 3 before you square. That is let x = 7+ or 10, then h(10)= 72=49. This will also be true with f(x) and h(x). For f(x) and h(x) to be equal then just let the x in h(x) be 3 more than the x in f(x) and you'll have equality(for the purest out there, I know there is another x value but that x value defeats my point). What does this do to the graph of h(x) when you compare it to the graph of f(x)??
 
You must log in or register to reply here.
Top