﻿ Graphing Tools: Vertical and Horizontal Scaling

# Graphing Tools: Vertical and Horizontal Scaling

You may want to review:

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: $$\cssId{s30}{\bigl(x, \overset{\text{y-value}}{\overbrace{ f(x)}} \bigr)}$$ 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:

$$\cssId{s36}{\bigl(x, \overset{\text{y-value}}{\overbrace{ f(x)}} \bigr)}$$
• 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} \cssId{s57}{\text{original equation:}} &\quad \cssId{s58}{y=f(x)}\cr \cssId{s59}{\text{new equation:}} &\quad \cssId{s60}{y=3f(x)} \end{align}

interpretation of new equation:

$$\cssId{s62}{\overset{\text{the new y-values}}{\overbrace{ \strut\ \ y\ \ }}} \cssId{s63}{\overset{\text{are}}{\overbrace{ \strut\ \ =\ \ }}} \cssId{s64}{\overset{\text{three times}\ \ \ }{\overbrace{ \strut \ \ 3\ \ }}} \cssId{s65}{\overset{\text{the previous y-values}}{\overbrace{ \strut\ \ f(x)\ \ }}}$$

## 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 $\,(a,kb)\,$ 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} \cssId{sb16}{\text{original equation:}} &\quad \cssId{sb17}{y=f(x)}\cr \cssId{sb18}{\text{new equation:}} &\quad \cssId{sb19}{y=f(3x)} \end{align}

interpretation of new equation:

$$\cssId{sb21}{y = f( \overset{\text{replace x by 3x}}{\overbrace{ \ \ 3x\ \ }}} )$$
• 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 $\,kx\,$ to give the new equation $\,y=f(kx)\,.$

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 $\,y=f(kx)\,.$

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 $\,(ka,b)\,$ 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  $\,y = f(kx)\,$  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 $\,(3a,b)\,.$