HORIZONTAL AND VERTICAL STRETCHING/SHRINKING

$y = f(x)$

$y = 2f(x)\,$
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
$y = f(x)$

$y = f(2x)\,$
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
vertical stretching/shrinking changes the $y$-values of points;
transformations that affect the $\,y\,$-values are intuitive
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 web exercise 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 on this current web page duplicate those in Graphing Tools: Vertical and Horizontal Scaling.

IDEAS REGARDING VERTICAL SCALING (STRETCHING/SHRINKING)
IDEAS REGARDING HORIZONTAL SCALING (STRETCHING/SHRINKING)
  • Points on the graph of [beautiful math coming... please be patient] $\,y=f(x)\,$ 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 $\,y=f(3x)\,$: [beautiful math coming... please be patient] $$ \begin{align} \text{original equation:} &\quad y=f(x)\cr\cr \text{new equation:} &\quad y=f(3x) \end{align} $$ [beautiful math coming... please be patient] $$ \begin{gather} \text{interpretation of new equation:}\cr\cr y = f( \overset{\text{replace x by 3x}}{\overbrace{ \ \ 3x\ \ }} ) \end{gather} $$ 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 [beautiful math coming... please be patient] $\,y=f(x)\,$.
    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 [beautiful math coming... please be patient] $\,y=f(k\,x)\,$.

    Additionally:
    Let $\,k\gt 1\,$.
    Start with the equation [beautiful math coming... please be patient] $\,y=f(x)\,$.
    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 [beautiful math coming... please be patient] $\,y=f(\frac{x}{k})\,$.

    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 [beautiful math coming... please be patient] $\,y\,$
(that is, transformations that change the $\,y$-values of the points), we say:

DO THIS to the previous $\,y$-value.

For transformations involving [beautiful math coming... please be patient] $\,x\,$
(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   [beautiful math coming... please be patient] $\,y=f(x)\,$   to   $\,y = kf(x)\,$   for   $\,k\gt 0$

horizontal scaling:
going from   [beautiful math coming... please be patient] $\,y = f(x)\,$   to   $\,y = f(k\,x)\,$   for   $\,k\gt 0$

Make sure you see the difference between (say) [beautiful math coming... please be patient] $\,y = 3f(x)\,$ and $\,y = f(3x)\,$!

In the case of [beautiful math coming... please be patient] $\,y = 3f(x)\,$, 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 [beautiful math coming... please be patient] $\,y = f(3x)\,$, 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.

When you're done practicing, move on to:
reflecting about axes, and the absolute value transformation


On this exercise, you will not key in your answer.
However, you can check to see if your answer is correct.
PROBLEM TYPES:
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AVAILABLE MASTERED IN PROGRESS

(MAX is 93; there are 93 different problem types.)