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!
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 $\,y=3f(x)\,$:
$$
\begin{align}
\cssId{s57}{\text{original equation:}} &\quad \cssId{s58}{y=f(x)}\cr\cr
\cssId{s59}{\text{new equation:}} &\quad \cssId{s60}{y=3f(x)}
\end{align}
$$
$$
\begin{gather}
\cssId{s61}{\text{interpretation of new equation:}}\cr\cr
\cssId{s62}{\overset{\text{the new y-values}}{\overbrace{
\strut\ \ y\ \ }}}
\cssId{s63}{\overset{\text{are}}{\overbrace{
\strut\ \ =\ \ }}}
\cssId{s64}{\overset{\quad\text{three times}\quad}{\overbrace{
\strut \ \ 3\ \ }}}
\cssId{s65}{\overset{\qquad\text{the previous y-values}\quad}{\overbrace{
\strut\ \ f(x)\ \ }}}
\end{gather}
$$
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).
LESSON READ-THROUGH (part 2 of 2)
by Dr. Carol JVF Burns (website creator)
Follow along with the highlighted text while you listen!
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 $\,y=f(3x)\,$:
$$
\begin{align}
\cssId{sb16}{\text{original equation:}} &\quad \cssId{sb17}{y=f(x)}\cr\cr
\cssId{sb18}{\text{new equation:}} &\quad \cssId{sb19}{y=f(3x)}
\end{align}
$$
$$
\begin{gather}
\cssId{sb20}{\text{interpretation of new equation:}}\cr\cr
\cssId{sb21}{y = f(
\overset{\text{replace x by 3x}}{\overbrace{
\ \ 3x\ \ }}}
)
\end{gather}
$$
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
by practicing the exercise at the bottom of this page.