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

## Feb 12, 2018

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

#### Explanation:

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

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

Feb 13, 2018

Here is a more detailed explanation

#### Explanation:

The problems are special cases of a more general problem :

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

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

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

This one is easy - all you have to do is note that if $\left(x , y\right)$ is a point on the first graph, then $\left(x , y + d\right)$ 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 $y = f \left(x\right) + 1$ will be As you can see, the graph for $y = f \left(x\right) + 1$ (the solid purple line) is obtained by simply pushing the graph for $y = f \left(x\right)$ (the gray dashed line) up by one unit.

The graph for $y = f \left(x\right) - 1$ can be found by pushing the original graph down by one unit : 2) Given the graph for $y = f \left(x\right)$, what is the graph of $y = f \left(x + c\right)$ ?

It is easy to see that if $\left(x , y\right)$ is a point on the $y = f \left(x\right)$ graph, then $\left(x - c , y\right)$ will be a point on the $y = f \left(x + c\right)$ graph. This means that you can get the graph of $y = f \left(x + c\right)$ from the graph of $y = f \left(x\right)$ 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 $y = f \left(x + 1\right)$ can be found by pushing the original graph to the left by one unit : while that for $y = f \left(x - 1\right)$ involves pushing the original graph to the right by one unit : 3) Given the graph for $y = f \left(x\right)$, what is the graph of $y = f \left(b x\right)$ ?

Since $f \left(x\right) = f \left(b \times \frac{x}{b}\right)$ it follows that if $\left(x , y\right)$ is a point on the $y = f \left(x\right)$ graph, then $\left(\frac{x}{b} , y\right)$ is a point on the $y = f \left(b x\right)$ 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 $\frac{1}{b}$ for the case where $0 < b < 1$

The graph for $y = f \left(2 x\right)$ is 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 $x = \frac{1}{2}$.

On the other hand, the graph for $y = f \left(\frac{x}{2}\right)$ is Note that this graph is twice as broad (squeezing by $\frac{1}{2}$ 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 $y = f \left(- x\right)$, and then
• squeeze the resulting graph by $| b |$

Note that for each point $\left(x , y\right)$ of the original graph, the point $\left(- x , y\right)$ is a point on the graph of $y = f \left(- x\right)$ - so the new graph can be found by reflecting the old one about the $Y$ axis. As an illustration of the two step process, consider the graph of $y = f \left(- 2 x\right)$ shown below : Here the original curve, that for $y = f \left(x\right)$ is first flipped about the $Y$ axis to get the curve for $y = f \left(- x\right)$ (the thin cyan line). This is then squeezed by a factor of $2$ to get the curve for $y = f \left(- 2 x\right)$ - the thick purple curve.

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

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 $\frac{1}{a}$ along the $Y$ axis.

The curve below is for $y = 2 f \left(x\right)$ 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 $0 < a < 1$ 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 $y = - f \left(x\right)$
2. Stretch the curve by $| a |$ along the $Y$ axis.

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

$y = a f \left(b x + c\right) + d$

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

1. Plot the curve of $y = f \left(x + c\right)$ : shift the graph by a distance $c$ to the left
2. Then plot that of $y = f \left(b x + c\right)$ : 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 $y = a f \left(b x + c\right)$ : 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 $\left(x , y\right)$ is a point on the $y = f \left(x\right)$ graph, then the point
$\left(\frac{x - c}{b} , a y + d\right)$ is on the $y = a f \left(b x + c\right) + d$ graph.

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

First - the shift to the left by 3 units Then : squeeze by a factor of 2 along the $X$ axis Then, flipping the graph over about the $X$ axis and then scaling by a factor of 2 along $Y$ Finally, shifting the curve up by 1 unit - and we are done! 