﻿ Graphing Tools: Vertical and Horizontal Scaling
GRAPHING TOOLS:
VERTICAL AND HORIZONTAL SCALING

There are things that you can DO to an equation of the form $\,y=f(x)\,$
that will change the graph in a variety of ways.

For example, you can move the graph up or down, left or right,
reflect about the $\,x\,$ or $\,y\,$ axes, stretch or shrink vertically or horizontally.

An understanding of these transformations makes it easy to graph a wide variety of functions,
by starting with a ‘basic model’ and then applying a sequence of transformations to change it to the desired function.

In this discussion, we will explore stretching and shrinking a graph, both vertically and horizontally.

When you finish studying this lesson, you should be able to do a problem like this:

GRAPH: $\,y=2{\text{e}}^{5x}\,$

• Start with the graph of $\,y={\text{e}}^x\,$.
(This is the ‘basic model’.)
• Multiply the previous $\,y$-values by $\,2\,$, giving the new equation $\,y=2{\text{e}}^x\,$.
This produces a vertical stretch, where the $\,y$-values on the graph get multiplied by $\,2\,$.
• Replace every $\,x\,$ by $\,5x\,$, giving the new equation $\,y=2{\text{e}}^{5x}\,$.
This produces a horizontal shrink, where the $\,x$-values on the graph get divided by $\,5\,$.

Here are ideas that are needed to understand graphical transformations.

IDEAS REGARDING FUNCTIONS AND THE GRAPH OF A FUNCTION

• A function is a rule:
it takes an input, and gives a unique output.
• If $\,x\,$ is the input to a function $\,f\,$,
then the unique output is called $\,f(x)\,$ (which is read as ‘$\,f\,$ of $\,x\,$’).
• The graph of a function is a picture of all of its (input,output) pairs.
We put the inputs along the horizontal axis (the $\,x\,$-axis),
and the outputs along the vertical axis (the $\,y\,$-axis).
• Thus, the graph of a function $\,f\,$ is a picture of all points of the form $\,\bigl(x, \overset{\text{y-value}}{\overbrace{ f(x)}} \bigr) \,$.
Here, $\,x\,$ is the input, and $\,f(x)\,$ is the corresponding output.
• The equation $\,y=f(x)\,$ is an equation in two variables, $\,x\,$ and $\,y\,$.
A solution is a choice for $\,x\,$ and a choice for $\,y\,$ that makes the equation true.
Of course, in order for this equation to be true, $\,y\,$ must equal $\,f(x)\,$.
Thus, solutions to the equation $\,y=f(x)\,$ are points of the form $\,\bigl(x, \overset{\text{y-value}}{\overbrace{ f(x)}} \bigr) \,$.
• Compare the previous two ideas!
To ‘graph the function $\,f\,$’ means to show all points of the form $\,\bigl(x,f(x)\bigr)\,$.
To ‘graph the equation $\,y=f(x)\,$’ means to show all points of the form $\,\bigl(x,f(x)\bigr)\,$.
These two requests mean exactly the same thing!

IDEAS REGARDING VERTICAL SCALING (STRETCHING/SHRINKING)
• Points on the graph of $\,y=f(x)\,$ are of the form $\,\bigl(x,f(x)\bigr)\,$.
Points on the graph of $\,y=3f(x)\,$ 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 makes the graph steeper.
• Points on the graph of $\,y=f(x)\,$ are of the form $\,\bigl(x,f(x)\bigr)\,$.
Points on the graph of $\,y=\frac13f(x)\,$ 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 makes 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 $\,y=f(x)\,$
and asked about the graph of $\,y=3f(x)\,$: \begin{align} \text{original equation:} &\quad y=f(x)\cr\cr \text{new equation:} &\quad y=3f(x) \end{align} $$\begin{gather} \text{interpretation of new equation:}\cr\cr \overset{\text{the new y-values}}{\overbrace{ \strut\ \ y\ \ }} \overset{\text{are}}{\overbrace{ \strut\ \ =\ \ }} \overset{\quad\text{three times}\quad}{\overbrace{ \strut \ \ 3\ \ }} \overset{\qquad\text{the previous y-values}\quad}{\overbrace{ \strut\ \ f(x)\ \ }} \end{gather}$$
• Summary of vertical scaling:

Let $\,k \gt 1\,$.
Start with the equation $\,y=f(x)\,$.
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 makes the graph steeper, and is called a vertical stretch.

Let $\,0 \lt k \lt 1\,$.
Start with the equation $\,y=f(x)\,$.
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 makes 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 $\,y=f(x)\,$ 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 counter-intuitive—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 $\,y=f(3x)\,$: \begin{align} \text{original equation:} &\quad y=f(x)\cr\cr \text{new equation:} &\quad y=f(3x) \end{align} $$\begin{gather} \text{interpretation of new equation:}\cr\cr y = f( \overset{\text{replace x by 3x}}{\overbrace{ \ \ 3x\ \ }} ) \end{gather}$$ 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 $\,y=f(x)\,$. 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 $\,y=f(k\,x)\,$. Additionally: Let $\,k\gt 1\,$. Start with the equation $\,y=f(x)\,$. 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 $\,y=f(\frac{x}{k})\,$. 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 $\,y\,$
(that is, transformations that change the $\,y$-values of the points), we say:

DO THIS to the previous $\,y$-value.

For transformations involving $\,x\,$
(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   $\,y=f(x)\,$   to   $\,y = kf(x)\,$   for   $\,k\gt 0$

horizontal scaling:
going from   $\,y = f(x)\,$   to   $\,y = f(k\,x)\,$   for   $\,k\gt 0$

Make sure you see the difference between (say) $\,y = 3f(x)\,$ and $\,y = f(3x)\,$!

In the case of $\,y = 3f(x)\,$, 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 $\,y = f(3x)\,$, 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 counter-intuitive.
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 counter-intuitive.
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 $\,(3a,b)\,$.
Master the ideas from this section