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

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

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Answer 1 · 2018-02-12T04:52:19

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

Explanation:

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

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

Answer 2 · 2018-02-13T09:38:27

Here is a more detailed explanation

Explanation:

The problems are special cases of a more general problem :

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

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

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.

enter image source here

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

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

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

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

enter image source here

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

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

enter image source here

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

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

enter image source here

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

enter image source here

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

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

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

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

enter image source here

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 $x = 1$ $x=1$ to $x = \frac{1}{2}$ $x=1/2$ .

On the other hand, the graph for $y = f (\frac{x}{2})$ $y=f(x/2)$ is

enter image source here

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

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

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

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

enter image source here

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

enter image source here

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

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

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

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

enter image source here

Note that the while the peak is at the same value of $x$ $x$ - its height has doubled to 2 from 1. Of course it is not the peak only that has been stretched - the $y$ $y$ 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 $0 < a < 1$ $0<a<1$

enter image source here

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

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

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

enter image source here

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

enter image source here

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

$y = a f (b x + c) + d$ $y = a f(bx+c)+d$

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

Plot the curve of $y = f (x + c)$ $y=f(x+c)$ : shift the graph by a distance $c$ $c$ to the left
Then plot that of $y = f (b x + c)$ $y = f(bx+c)$ : squeeze the curve that you get from step 1 in the $X$ $X$ direction by the factor $| b |$ $|b|$ , (first flipping it about the $Y$ $Y$ axis if $b < 0$ $b<0$ )
Then plot the graph of $y = a f (b x + c)$ $y=af(bx+c)$ : scale the curve that you got from step 2 to by a factor of $a$ $a$ in the vertical direction.
Finally push the curve that you obtain in step 3 up by a distance $d$ $d$ 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 $(x, y)$ $(x,y)$ is a point on the $y = f (x)$ $y=f(x)$ graph, then the point
$(\frac{x - c}{b}, a y + d)$ $({x-c}/b,ay+d)$ is on the $y = a f (b x + c) + d$ $y=af(bx+c)+d$ graph.

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

First - the shift to the left by 3 units

enter image source here

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

enter image source here

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

enter image source here

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

enter image source here

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

2 Answers

Explanation:

Explanation:

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

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Explanation:

Explanation:

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