Graphing Tools: Vertical and Horizontal Translations

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 moving a graph up/down (vertical translations)
and moving a graph left/right (horizontal translations).

When you finish studying this lesson, you should be able to do a problem like this:

GRAPH: $\,y=(x-3)^2+5\,$

Start with the graph of $\,y=x^2\,$.
(This is the ‘basic model’.)
Add $\,5\,$ to the previous $\,y\,$-values, giving the new equation $\,y=x^2+5\,$.
This moves the graph UP $\,5\,$ units.
Replace every $\,x\,$ by $\,x-3\,$, giving the new equation $\,y=(x-3)^2+5\,$.
This moves the graph to the RIGHT $\,3\,$ units.

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 $\,\bigl(x, \overset{\text{y-value}}{\overbrace{ f(x)}} \bigr) \,$.
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 $\,\bigl(x, \overset{\text{y-value}}{\overbrace{ f(x)}} \bigr) \,$.
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 TRANSLATIONS (MOVING UP/DOWN)

Points on the graph of $\,y=f(x)\,$ are of the form $\,\bigl(x,f(x)\bigr)\,$.
Points on the graph of $\,y=f(x)+3\,$ are of the form $\,\bigl(x,f(x)+3\bigr)\,$.
Thus, the graph of $\,y=f(x)+3\,$ is the same as the graph of $\,y=f(x)\,$, shifted UP three units.
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 $\,y=f(x)+3\,$: $$ \begin{align} \text{original equation:} &\quad y=f(x)\cr\cr \text{new equation:} &\quad y=f(x) + 3 \end{align} $$ $$ \begin{gather} \text{interpretation of new equation:}\cr\cr \overset{\text{the new y-values}}{\overbrace{ \strut\ \ y\ \ }} \overset{\text{are}}{\overbrace{ \strut\ \ =\ \ }} \overset{\quad\text{the previous y-values}\quad}{\overbrace{ \strut f(x)}} \overset{\qquad\text{with 3 added to them}\quad}{\overbrace{ \strut\ \ + 3\ \ }} \end{gather} $$
Summary of vertical translations:
Let $\,p\,$ be a positive number.

Start with the equation $\,y=f(x)\,$.
Adding $\,p\,$ to the previous $\,y\,$-values gives the new equation $\,y=f(x)+p\,$.
This shifts the graph UP $\,p\,$ units.
A point $\,(a,b)\,$ on the graph of $\,y=f(x)\,$ moves to a point $\,(a,b+p)\,$ on the graph of $\,y=f(x)+p\,$.

Additionally:
Start with the equation $\,y=f(x)\,$.
Subtracting $\,p\,$ from the previous $\,y\,$-values gives the new equation $\,y=f(x)-p\,$.
This shifts the graph DOWN $\,p\,$ units.
A point $\,(a,b)\,$ on the graph of $\,y=f(x)\,$ moves to a point $\,(a,b-p)\,$ on the graph of $\,y=f(x)-p\,$.

This transformation type (shifting up and down) is formally called vertical translation.

IDEAS REGARDING HORIZONTAL TRANSLATIONS (MOVING LEFT/RIGHT)

Points on the graph of $\,y=f(x)\,$ are of the form $\,\bigl(x,f(x)\bigr)\,$.
Points on the graph of $\,y=f(x+3)\,$ are of the form $\,\bigl(x,f(x+3)\bigr)\,$.
How can we locate these desired points $\,\bigl(x,f(x+3)\bigr)\,$?

First, go to the point $\,\bigl(x+3\,,\,f(x+3)\bigr)\,$ on the graph of $\,y=f(x)\,$.
This point has the $\,y$-value that we want, but it has the wrong $\,x$-value.

Move this point $\,3\,$ units to the left.
Thus, the $\,y$-value stays the same, but the $\,x$-value is decreased by $\,3\,$.
This gives the desired point $\,\bigl(x,f(x+3)\bigr)\,$.

Thus, the graph of $\,y=f(x+3)\,$ is the same as the graph of $\,y=f(x)\,$, shifted LEFT three units.

Thus, replacing $\,x\,$ by $\,x+3\,$ moved the graph LEFT (not right, as might have been expected!)
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 $\,y=f(x+3)\,$: $$ \begin{align} \text{original equation:} &\quad y=f(x)\cr\cr \text{new equation:} &\quad y=f(x+3) \end{align} $$ $$ \begin{gather} \text{interpretation of new equation:}\cr\cr y = f( \overset{\text{replace x by x+3}}{\overbrace{ x+3}} ) \end{gather} $$ Replacing every $\,x\,$ by $\,x+3\,$ in an equation moves the graph $\,3\,$ units TO THE LEFT.
Summary of horizontal translations:
Let $\,p\,$ be a positive number.

Start with the equation $\,y=f(x)\,$.
Replace every $\,x\,$ by $\,x+p\,$ to give the new equation $\,y=f(x+p)\,$.
This shifts the graph LEFT $\,p\,$ units.
A point $\,(a,b)\,$ on the graph of $\,y=f(x)\,$ moves to a point $\,(a-p,b)\,$ on the graph of $\,y=f(x+p)\,$.

Additionally:
Start with the equation $\,y=f(x)\,$.
Replace every $\,x\,$ by $\,x-p\,$ to give the new equation $\,y=f(x-p)\,$.
This shifts the graph RIGHT $\,p\,$ units.
A point $\,(a,b)\,$ on the graph of $\,y=f(x)\,$ moves to a point $\,(a+p,b)\,$ on the graph of $\,y=f(x-p)\,$.

This transformation type (shifting left and right) is formally called horizontal translation.

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 translations:
going from $\,y=f(x)\,$ to $\,y = f(x) \pm c\,$

horizontal translations:
going from $\,y = f(x)\,$ to $\,y = f(x\pm c)\,$

Make sure you see the difference between (say) $\,y = f(x) + 3\,$ and $\,y = f(x+3)\,$!

In the case of $\,y = f(x) + 3\,$, the $\,3\,$ is ‘on the outside’;
we're dropping $\,x\,$ in the $\,f\,$ box, getting the corresponding output, and then adding $\,3\,$ to it.
This is a vertical translation.

In the case of $\,y = f(x + 3)\,$, the $\,3\,$ is ‘on the inside’;
we're adding $\,3\,$ to $\,x\,$ before dropping it into the $\,f\,$ box.
This is a horizontal translation.

EXAMPLES:

Question:
Start with $\,y = f(x)\,$.
Move the graph TO THE RIGHT $\,2\,$.
What is the new equation?

Solution:
This is a transformation involving $\,x\,$; it is counter-intuitive.
You must replace every $\,x\,$ by $\,x-2\,$.
The new equation is:
$\,y = f(x-2)\,$

Question:
Start with $\,y = x^2\,$.
Move the graph DOWN $\,3\,$.
What is the new equation?

Solution:
This is a transformation involving $\,y\,$; it is intuitive.
You must subtract $\,3\,$ from the previous $\,y\,$-value.
The new equation is:
$\,y = x^2 - 3\,$

Question:
Let $\,(a,b)\,$ be a point on the graph of $\,y = f(x)\,$.
Then, what point is on the graph of $\,y = f(x+5)\,$?

Solution:
This is a transformation involving $\,x\,$; it is counter-intuitive.
Replacing every $\,x\,$ by $\,x+5\,$ in an equation causes the graph to shift $\,5\,$ units to the LEFT.
Thus, the new point is $\,(a-5,b)\,$.

Master the ideas from this section
by practicing the exercise at the bottom of this page.

When you're done practicing, move on to:
Vertical and Horizontal Scaling
(stretching and shrinking)

(MAX is 64; there are 64 different problem types.)