audio read-through Graphing Tools: Vertical and Horizontal Scaling (Part 2)

(This page is Part 2. Click here for Part 1.)

Ideas Regarding Horizontal Scaling (Stretching/Shrinking)

$$ \begin{align} \cssId{s16}{\text{original equation:}} &\quad \cssId{s17}{y=f(x)}\cr \cssId{s18}{\text{new equation:}} &\quad \cssId{s19}{y=f(3x)} \end{align} $$

interpretation of new equation:

$$ \cssId{s21}{y = f( \overset{\text{replace $x$ by $3x$}}{\overbrace{ \ \ 3x\ \ }}} ) $$

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