#y=f(x)# is given. Graph, #y=f(3x)-2# and #y=-f(x-1)#?

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2 Answers
Feb 12, 2018

Don't have graph paper handy - so I hope that the description helps!

Explanation:

For #y=f(3x)-2# first squeeze the given graph along the #x# axis by a factor of 3 (so that the left hand minimum, say, occurs at #x=-2/3# ), and then push the whole graph down by 2 units. Thus the new graph will have a minimum at #x = -2/3# with a value of #y= -2#, a maximum at #(0,0)# and another minimum at #(4/3, -4)#

For #y=-f(x-1)# first shift the graph 1 unit to the right , then flip it upside down! So, the new graph will ave two maxima at #(-1,0)# and #(5,2)# and a minimum at #(1,-2) #

Feb 13, 2018

Here is a more detailed explanation

Explanation:

The problems are special cases of a more general problem :

Given the graph for #y=f(x)#, what is the graph of #y = a f(b x+c)+d# ?

(the first one is for #a=1, b= 3, c=0, d=-2#, while the second one is for #a=-1, b=1,c=-1, d=0# )

I will try to explain the answer in steps, by tackling the problem one step at a time. It will be a pretty long answer - but hopefully the general principle will be clear by the end of it.

For illustration I'll use a particular curve that I am showing below, but the idea will work in general.

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(If anyone is interested, the function that is being plotted here is #f(x) = exp(-{(x-1)^2}/2)#

1) Given the graph for #y=f(x)#, what is the graph of #y = f( x)+d# ?

This one is easy - all you have to do is note that if #(x,y)# is a point on the first graph, then #(x,y+d)# is a point on the second. This means that the second graph is higher than the first by a distance #d# (of course, if #d# is negative, it is lower than the first graph by #|d|# ).

So, the graph of #y=f(x)+1# will be

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As you can see, the graph for #y = f(x)+1# (the solid purple line) is obtained by simply pushing the graph for #y=f(x)# (the gray dashed line) up by one unit.

The graph for #y=f(x)-1# can be found by pushing the original graph down by one unit :

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2) Given the graph for #y=f(x)#, what is the graph of #y = f( x+c)# ?

It is easy to see that if #(x,y)# is a point on the #y=f(x)# graph, then #(x-c,y)# will be a point on the #y = f(x+c)# graph. This means that you can get the graph of #y = f( x+c)# from the graph of #y = f( x)# simply by shifting it to the left by #c# (of course, if #c# is negative, you must shift the original graph by #|c|# to the right.
As an example, the graph for #y=f(x+1)# can be found by pushing the original graph to the left by one unit :

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while that for #y=f(x-1)# involves pushing the original graph to the right by one unit :

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3) Given the graph for #y=f(x)#, what is the graph of #y = f( bx)# ?

Since #f(x) = f(b times x/b)# it follows that if #(x,y)# is a point on the #y = f(x)# graph, then #(x/b , y)# is a point on the #y=f(bx)# graph.

This means that the original graph has to be squeezed by a factor of #b# along the #x# axis. Of course, the squeezing by #b# is really a stretching by #1/b# for the case where #0 < b <1#

The graph for #y=f(2x)# is

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Note that the while the height stays the same at 1, the width shrinks by a factor of 2. In particular, the peak of the original curve has shifted from #x=1# to #x=1/2#.

On the other hand, the graph for #y=f(x/2)# is

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Note that this graph is twice as broad (squeezing by #1/2# being the same as stretching by a factor of 2), and the peak has also moved from #x=1# to #x=2#.

A special mention must be made of the case where #b# is negative. It is best perhaps to then think of this as a two-step process

  • First find the graph of #y=f(-x)#, and then
  • squeeze the resulting graph by #|b|#

Note that for each point #(x,y)# of the original graph, the point #(-x,y)# is a point on the graph of #y=f(-x)# - so the new graph can be found by reflecting the old one about the #Y# axis.

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As an illustration of the two step process, consider the graph of #y=f(-2x)# shown below :

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Here the original curve, that for #y=f(x)# is first flipped about the #Y# axis to get the curve for #y=f(-x)# (the thin cyan line). This is then squeezed by a factor of #2# to get the curve for #y=f(-2x)# - the thick purple curve.

4) Given the graph for #y=f(x)#, what is the graph of #y = af( x)# ?
The pattern is the same here - if #(x,y)# is a point on the original curve then #(x,ay)# is a point on the graph of #y=af(x)#

This means that for a positive #a#, the graph gets stretched by a factor of #a# along the #Y# axis. Again, a value of #a# between 0 and 1 means that instead of being stretched, the curve will actually be squeezed by a factor of #1/a# along the #Y# axis.

The curve below is for # y = 2f(x)#

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Note that the while the peak is at the same value of #x# - its height has doubled to 2 from 1. Of course it is not the peak only that has been stretched - the #y# coordinate of every point of the original curve has been doubled to get the new curve.

The figure below illustrates the squeezing that occurs when #0<a<1#

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Once again, the case for #a<0# takes special care - and it is better if you do this in two steps

  1. First flip the curve upside down about the #X# axis to get the curve for #y=-f(x)#
  2. Stretch the curve by #|a|# along the #Y# axis.

The curve for #y=-f(x)# is

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while the picture below illustrates the two steps involved in drawing the curve for #y = -2f(x)#

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Putting it all together

Now that we have gone through the individual steps, let us put them all together! The procedure for drawing the curve for

# y = a f(bx+c)+d#

starting from that of #y=f(x)# is essentially composed of the following steps

  1. Plot the curve of #y=f(x+c)# : shift the graph by a distance #c# to the left
  2. Then plot that of #y = f(bx+c)# : squeeze the curve that you get from step 1 in the #X# direction by the factor #|b|#, (first flipping it about the #Y# axis if #b<0#)
  3. Then plot the graph of #y=af(bx+c)# : scale the curve that you got from step 2 to by a factor of #a# in the vertical direction.
  4. Finally push the curve that you obtain in step 3 up by a distance #d# to get the final result.

Of course you need to carry out all four steps only in extreme cases - often a smaller number of steps will do! Also, the sequence of steps is important.
In case you are wondering, these steps follow from the fact that if #(x,y)# is a point on the #y=f(x)# graph, then the point
# ({x-c}/b,ay+d)# is on the #y=af(bx+c)+d# graph.

Let me illustrate the process by an example with our function #f(x)#. Let us try to construct the graph for #y = -2f(2x+3)+1#

First - the shift to the left by 3 units

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Then : squeeze by a factor of 2 along the #X# axis

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Then, flipping the graph over about the #X# axis and then scaling by a factor of 2 along #Y#

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Finally, shifting the curve up by 1 unit - and we are done!

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