audio read-through Horizontal and Vertical Stretching/Shrinking

 

vertical stretch vertical shrink
$y = f(x)$

$y = 2f(x)\,$
Vertical stretch;
$y$-values are doubled;
points get farther away from $x$-axis
$y = f(x)$

$y = \frac{f(x)}{2}\,$
Vertical shrink;
$y$-values are halved;
points get closer to $x$-axis

Vertical stretching/shrinking changes the $y$-values of points.

Transformations that affect the $y$-values are intuitive.

horizontal shrink horizontal stretch
$y = f(x)$

$y = f(2x)\,$
Horizontal shrink;
$x$-values are halved;
points get closer to $y$-axis
$y = f(x)$

$y = f(\frac x2)$
Horizontal stretch;
$x$-values are doubled;
points get farther away from $y$-axis

Horizontal stretching/shrinking changes the $x$-values of points.

Transformations that affect the $x$-values are counter-intuitive.

Vertical/horizontal stretching/shrinking usually changes the shape of a graph.

The lesson Graphing Tools: Vertical and Horizontal Scaling in the Algebra II curriculum gives a thorough discussion of horizontal and vertical stretching and shrinking. The key concepts are repeated here.

The exercises in this lesson duplicate those in Graphing Tools: Vertical and Horizontal Scaling .

Ideas Regarding Vertical Scaling (Stretching/Shrinking)

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

Interpretation of new equation:

$$ \begin{gather} \overset{\text{the new y-values}}{\overbrace{ \strut\ \ y\ \ }} \overset{\text{are}}{\overbrace{ \strut\ \ =\ \ }} \overset{\text{three times}\ \ }{\overbrace{ \strut \ \ 3\ \ }} \overset{\text{the previous y-values}}{\overbrace{ \strut\ \ f(x)\ \ }} \end{gather} $$

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 tends to make 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 tends to make 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)

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