Explanation for the rules around modifying graphs.
Overall rule: If x or y in the equation y = f(x) is replaced by a modification,
the effect on the graph is that
1.) if the change is made to x (or y), the graph changes horizontally (or
vertically)
2) the change in the graph is the opposite of the change suggested by the
formula
That is (in gruesome detail): Suppose you have the graph of y = f(x), then
the graph of . . .
y = f(x+k)
y = f(x-k)
y = f(cx)
y = f(x/c)
y+k = f(x)
y-k = f(x)
cy = f(x)
y/c = f(x)
is just the graph of y=f(x) . . .
moved left k units
moved right k units
contracted horizontally by a factor of c
stretched horizontally by a factor of c
moved down k units
moved up k units
contracted vertically by a factor of c
stretched vertically by a factor of c
Why?
An example best illustrates: Suppose we have the graph of y=x2 and we
want the graph of y=(x+2)2. For y=x2 we actually have all the points, since
we know the graph, but let’s work with just the ones in the following T-chart
so we can examine the problem numerically.
x
-2
-1
0
1
2
y=x2
4
1
0
1
4
Note that our formula y=(x+2)2 produces all the squares, by choosing
carefully which x we will use. For instance, if we wish to get y=4 by
squaring –2, we have to use x = -4. (since ((-4)+2)2 = (-2)2 = 4). Thus we
have (-4,4) in the T-chart for y=(x+2)2. In a similar way, we have the other
points in the chart:
x
-4
-3
-2
-1
0
y=(x+2)2
4
1
0
1
4
Note that each of the new x values is just the old x-value reduced by 2. The
reason is that we had to reduce it by 2 in order that, when the formula adds
2, we have the correct input for the squaring function. This argument
applies to all x-values in the domain of the original function. The result is
that the graph of the new function has exactly the same points as the old
function, except that all the x-values are reduced by 2. That is, the entire
graph moves 2 units to the left.
Note that in this example, the modification was to replace x by x+2. Our
rule says that we should modify the graph horizontally and that we should
reduce each x by 2 (the opposite of adding 2.) That is exactly what we came
up with by working through it step by step, but now we understand why the
rule works -- at least for this example.
You might be able to see that the reasoning we just did could be modified in
predictable ways if
1. we added 7 to x instead of 2. (move 7 units left)
2. we subtracted 3 from x. (move 3 units right)
3. we added 4 to y (move 4 units down)
4. we subtracted .5 units from y (move .5 units up)
5. we multiplied x by 4 (divide each x by 4 – contract horizontally by a
factor of 4)
6. we divided each y by 2 (multiply each y by 2 – stretch vertically by a
factor of 2)
If you don’t understand any of these in ordinary common sense, use the
paragraphs on x+2 as a model and see if you can rewrite the argument for
the problem you are working.
Test yourself by quickly finding the graphs of
1.
2.
3.
4.
5.
y = (3x)2
y-8 = x2
y = x-5 (use the new method)
y = 7x (use the new method)
y/7 = x2
Other variations:
1. Note that you may have to use some of the usual language tricks to do
a particular problem a particular way. For instance, if you have the
graph of y=f(x) and want to know how far to the right to shift for the
graph of y = f(x+2), you know that you actually shift 2 units to the
left, but to say it in terms of a shift to the right, you could say shift –2
units to the right. Similarly to stretch by a factor of 1/5 just contract
by a factor of 5.
2. You may need to string several of the basic steps together to answer a
more complicated question. For instance, if you have the graph of
y=f(x) and want the graph of y=f(x/3+7), you could do it in two steps:
shift eh graph of f(x) 7 units left to get the graph of y=f(x+7). Next,
contract that result by a factor of 3 horizontally to get the graph of
y=f(x/3+7).
Try these: find the graphs of
1.
2.
3.
4.
y = (2x+3)2
y = (2(x+3))2
(y-2)/5 = x2
y/5-2 = x2