
Points on the graph of
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$\,y=f(x)\,$ are of the form $\,\bigl(x,f(x)\bigr)\,$.
Points on the graph of
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$\,y=3f(x)\,$ are of the form $\,\bigl(x,3f(x)\bigr)\,$.
Thus, the graph of $\,y=3f(x)\,$ is found by taking the graph of $\,y=f(x)\,$,
and multiplying the $\,y$values by $\,3\,$.
This moves the points farther from the $\,x$axis, which tends to make the graph steeper.

Points on the graph of
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$\,y=f(x)\,$ are of the form $\,\bigl(x,f(x)\bigr)\,$.
Points on the graph of
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$\,y=\frac13f(x)\,$ are of the form $\,\bigl(x,\frac13f(x)\bigr)\,$.
Thus, the graph of $\,y=\frac13f(x)\,$ is found by taking the graph of $\,y=f(x)\,$,
and multiplying the $\,y$values by $\,\frac13\,$.
This moves the points closer to the $\,x$axis, which tends to make the graph flatter.

Transformations involving $\,y\,$ work the way you would expect them to work—they are intuitive.

Here is the thought process you should use when you are given the graph of
[beautiful math coming... please be patient]
$\,y=f(x)\,$
and asked about the graph of
$\,y=3f(x)\,$:
[beautiful math coming... please be patient]
$$
\begin{align}
\text{original equation:} &\quad y=f(x)\cr\cr
\text{new equation:} &\quad y=3f(x)
\end{align}
$$
[beautiful math coming... please be patient]
$$
\begin{gather}
\text{interpretation of new equation:}\cr\cr
\overset{\text{the new yvalues}}{\overbrace{
\strut\ \ y\ \ }}
\overset{\text{are}}{\overbrace{
\strut\ \ =\ \ }}
\overset{\quad\text{three times}\quad}{\overbrace{
\strut \ \ 3\ \ }}
\overset{\qquad\text{the previous yvalues}\quad}{\overbrace{
\strut\ \ f(x)\ \ }}
\end{gather}
$$

Summary of vertical scaling:
Let $\,k \gt 1\,$.
Start with the equation
[beautiful math coming... please be patient]
$\,y=f(x)\,$.
Multiply the previous $\,y\,$values by $\,k\,$, giving the new equation
$\,y=kf(x)\,$.
The $\,y$values are being multiplied by a number greater than $\,1\,$, so they move farther from the $\,x$axis.
This tends to make the graph steeper, and is called a vertical stretch.
Let $\,0 \lt k \lt 1\,$.
Start with the equation
[beautiful math coming... please be patient]
$\,y=f(x)\,$.
Multiply the previous $\,y\,$values by $\,k\,$, giving the new equation
$\,y=kf(x)\,$.
The $\,y$values are being multiplied by a number between $\,0\,$ and $\,1\,$, so they move closer to the $\,x$axis.
This tends to make the graph flatter, and is called a vertical shrink.
In both cases, a point $\,(a,b)\,$ on the graph of $\,y=f(x)\,$ moves to a point $\,(a,k\,b)\,$
on the graph of $\,y=kf(x)\,$.
This transformation type is formally called vertical scaling (stretching/shrinking).
IDEAS REGARDING HORIZONTAL SCALING (STRETCHING/SHRINKING)

Points on the graph of
[beautiful math coming... please be patient]
$\,y=f(x)\,$ are of the form $\,\bigl(x,f(x)\bigr)\,$.
Points on the graph of $\,y=f(3x)\,$ are of the form $\,\bigl(x,f(3x)\bigr)\,$.

How can we locate these desired points $\,\bigl(x,f(3x)\bigr)\,$?
First, go to the point
$\,\color{red}{\bigl(3x\,,\,f(3x)\bigr)}\,$
on the graph of
$\,\color{red}{y=f(x)}\,$.
This point has the $\,y$value that we want, but it has the wrong $\,x$value.
The $\,x$value of this point is $\,3x\,$, but the desired $\,x$value is just $\,x\,$.
Thus, the current
$\,\color{purple}{x}$value must be divided by
$\,\color{purple}{3}\,$; the $\,\color{purple}{y}$value remains the same.
This gives the desired point
$\,\color{green}{\bigl(x,f(3x)\bigr)}\,$.
Thus, the graph of $\,y=f(3x)\,$ is the same as the graph of $\,y=f(x)\,$,
except that the $\,x$values have been divided by $\,3\,$ (not multiplied by $\,3\,$, which you might expect).
Notice that dividing the $\,x$values by $\,3\,$ moves them closer to the $\,y$axis; this is called a horizontal shrink.

Transformations involving $\,x\,$
do NOT work the way you would expect them to work!
They are counterintuitive—they are against your intuition.

Here is the thought process you should use when you are given the graph of $\,y=f(x)\,$
and asked about the graph of $\,y=f(3x)\,$:
[beautiful math coming... please be patient]
$$
\begin{align}
\text{original equation:} &\quad y=f(x)\cr\cr
\text{new equation:} &\quad y=f(3x)
\end{align}
$$
[beautiful math coming... please be patient]
$$
\begin{gather}
\text{interpretation of new equation:}\cr\cr
y = f(
\overset{\text{replace x by 3x}}{\overbrace{
\ \ 3x\ \ }}
)
\end{gather}
$$
Replacing every $\,x\,$ by
$\,3x\,$ in an equation
causes the $\,x$values in the graph to be DIVIDED by $\,3$.

Summary of horizontal scaling:
Let $\,k\gt 1\,$.
Start with the equation
[beautiful math coming... please be patient]
$\,y=f(x)\,$.
Replace every $\,x\,$ by $\,k\,x\,$ to
give the new equation $\,y=f(k\,x)\,$.
This causes the $\,x$values on the graph to be DIVIDED by $\,k\,$, which moves the points closer to the $\,y$axis.
This is called a horizontal shrink.
A point $\,(a,b)\,$ on the graph of $\,y=f(x)\,$ moves to a point $\,(\frac{a}{k},b)\,$ on the graph of
[beautiful math coming... please be patient]
$\,y=f(k\,x)\,$.
Additionally:
Let $\,k\gt 1\,$.
Start with the equation
[beautiful math coming... please be patient]
$\,y=f(x)\,$.
Replace every $\,x\,$ by $\,\frac{x}{k}\,$ to
give the new equation $\,y=f(\frac{x}{k})\,$.
This causes the $\,x$values on the graph to be MULTIPLIED by $\,k\,$, which moves the points farther away from the $\,y$axis.
This is called a horizontal stretch.
A point $\,(a,b)\,$ on the graph of $\,y=f(x)\,$ moves to a point $\,(k\,a,b)\,$ on the graph of
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$\,y=f(\frac{x}{k})\,$.
This transformation type is formally called horizontal scaling (stretching/shrinking).


DIFFERENT WORDS USED TO TALK ABOUT TRANSFORMATIONS INVOLVING $\,y\,$ and $\,x\,$
Notice that different words are used when talking about transformations involving
$\,y\,$, and transformations involving $\,x\,$.
For transformations involving
[beautiful math coming... please be patient]
$\,y\,$
(that is, transformations that change the $\,y$values of the points),
we say:
DO THIS to the previous $\,y$value.
For transformations involving [beautiful math coming... please be patient]
$\,x\,$
(that is, transformations that change the $\,x$values of the points),
we say:
REPLACE the previous $\,x$values by $\ldots$
MAKE SURE YOU SEE THE DIFFERENCE!
vertical scaling:
going from
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$\,y=f(x)\,$
to
$\,y = kf(x)\,$ for $\,k\gt 0$
horizontal scaling:
going from
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$\,y = f(x)\,$
to
$\,y = f(k\,x)\,$ for $\,k\gt 0$
Make sure you see the difference between (say)
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$\,y = 3f(x)\,$
and
$\,y = f(3x)\,$!
In the case of
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$\,y = 3f(x)\,$, the $\,3\,$ is ‘on the outside’;
we're dropping $\,x\,$ in the $\,f\,$ box, getting the corresponding output, and then multiplying by $\,3\,$.
This is a vertical stretch.
In the case of
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$\,y = f(3x)\,$, the $\,3\,$ is ‘on the inside’;
we're multiplying $\,x\,$ by $\,3\,$ before dropping it into the $\,f\,$ box.
This is a horizontal shrink.
EXAMPLES:
Question:
Start with $\,y = f(x)\,$.
Do a vertical stretch; the $\,y$values on the graph should be multiplied by $\,2\,$.
What is the new equation?
Solution:
This is a transformation involving $\,y\,$; it is intuitive.
You must multiply the previous $\,y$values by $\,2\,$.
The new equation is:
$\,y = 2f(x)\,$
Question:
Start with $\,y = f(x)\,$.
Do a horizontal stretch; the $\,x$values on the graph should get multiplied by $\,2\,$.
What is the new equation?
Solution:
This is a transformation involving $\,x\,$; it is counterintuitive.
You must replace every $\,x\,$ in the equation by $\,\frac{x}{2}\,$.
The new equation is:
$\,y = f(\frac{x}{2})\,$
Question:
Start with $\,y = x^3\,$.
Do a vertical shrink, where $\,(a,b) \mapsto (a,\frac{b}{4})\,$.
What is the new equation?
Solution:
This is a transformation involving $\,y\,$; it is intuitive.
You must multiply the previous $\,y$values by $\frac 14\,$.
The new equation is:
$\,y = \frac14 x^3\,$
Question:
Suppose $\,(a,b)\,$ is a point on the graph of $\,y = f(x)\,$.
Then, what point is on the graph of $\,y = f(\frac{x}{3})\,$?
Solution:
This is a transformation involving $\,x\,$; it is counterintuitive.
Replacing every $\,x\,$ by $\,\frac{x}{3}\,$ in the equation causes the $\,x$values on the graph to be multiplied by $\,3\,$.
Thus, the new point is $\,(3a,b)\,$.