audio read-through Graphing Tools: Vertical and Horizontal Scaling

Click here for a printable version of the discussion below.

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

Here are ideas that are needed to understand graphical transformations.

Ideas Regarding Functions and the Graph of a Function

Ideas Regarding Vertical Scaling (Stretching/Shrinking)

$$ \begin{align} \cssId{s57}{\text{original equation:}} &\quad \cssId{s58}{y=f(x)}\cr \cssId{s59}{\text{new equation:}} &\quad \cssId{s60}{y=3f(x)} \end{align} $$

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).

Read-Through, Part 2

Ideas Regarding Horizontal Scaling (Stretching/Shrinking)

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

interpretation of new equation:

$$ \cssId{sb21}{y = f( \overset{\text{replace $x$ by $3x$}}{\overbrace{ \ \ 3x\ \ }}} ) $$

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 $\,(3a,b)\,.$

Concept Practice