audio read-through Graphing Tools: Vertical and Horizontal Translations

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

Here are ideas that are needed to understand graphical transformations.

Ideas Regarding Functions and the Graph of a Function

Ideas Regarding Vertical Translations (Moving Up/Down)

$$ \begin{align} \cssId{s55}{\text{original equation:}} &\quad \cssId{s56}{y=f(x)}\cr \cssId{s57}{\text{new equation:}} &\quad \cssId{s58}{y=f(x) + 3} \end{align} $$

interpretation of new equation:

$$ \cssId{s60}{\overset{\text{the new y-values}}{\overbrace{ \strut\ \ y\ \ }}} \cssId{s61}{\overset{\text{are}}{\overbrace{ \strut\ \ =\ \ }}} \cssId{s62}{\overset{\text{the previous y-values}}{\overbrace{ \strut f(x)}}} \cssId{s63}{\ \ \overset{\text{with 3 added to them}}{\overbrace{ \strut\ \ + 3\ \ }}} $$

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.

Read-Through, Part 2

Ideas Regarding Horizontal Translations (Moving Left/Right)

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

interpretation of new equation:

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

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 $\,(a-5,b)\,.$

Concept Practice

Note: There are lots of questions like this:

‘Start with $\,y = f(x)\,.$ Move UP $\,2\,.$ What is the new equation?’

Here is the same question, stated more precisely:

‘Start with the graph of $\,y = f(x)\,.$ Move this graph UP $\,2\,.$ What is the equation of the new graph?’

All the ‘graphs’ are implicit in the problem statements. (When something is implicit then it's understood to be there, even though you can't see it.)