audio read-through Shifting Graphs Up/Down Left/Right

 

shifting a graph UP shifting a graph DOWN
$y = f(x)$

$y = f(x) + 2$
Up $2$
$y = f(x)$

$y = f(x) - 2$
Down $2$

Movements up and down change the $y$-values of points.

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

For example, to move up $\,2\,,$ you add $\,2\,$ to the previous $y$-value.

shifting a graph LEFT shifting a graph RIGHT
$y = f(x)$

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

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

Movements left and right change the $x$-values of points.

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

For example, to move left $\,2\,,$ you replace every $\,x\,$ by $\,x+2\,,$ not $\,x-2\,.$

Shifting up/down left/right does not change the shape of a graph.

The lesson Graphing Tools: Vertical and Horizontal Translations in the Algebra II curriculum gives a thorough discussion of shifting graphs up/down left/right. The key concepts are repeated here.

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

Ideas Regarding Vertical Translations (Moving Up/Down)
$$ \begin{align} \text{original equation:} &\quad y=f(x)\cr \text{new equation:} &\quad y=f(x) + 3 \end{align} $$

Interpretation of new equation:

$$ \begin{gather} \overset{\text{the new y-values}}{\overbrace{ \strut\ \ y\ \ }} \overset{\text{are}}{\overbrace{ \strut\ \ =\ \ }} \overset{\text{the previous y-values}}{\overbrace{ \strut f(x)}} \overset{\ \ \text{with 3 added to them}}{\overbrace{ \strut\ \ + 3\ \ }} \end{gather} $$

Summary of Vertical Translations

Let $\,p\,$ be a positive number.

Start with the equation $\,\color{purple}{y=f(x)}\,.$ Adding $\,\color{green}{p}\,$ to the previous $\color{green}{y}$-values gives the new equation $\,\color{green}{y=f(x)+p}\,.$ This shifts the graph UP $\,\color{green}{p}\,$ units.

shifting a graph UP

A point $\,\color{purple}{(a,b)}\,$ on the graph of $\,\color{purple}{y=f(x)}\,$ moves to a point $\,\color{green}{(a,b+p)}\,$ on the graph of $\,\color{green}{y=f(x)+p}\,.$

Additionally:

Start with the equation $\,\color{purple}{y=f(x)}\,.$ Subtracting $\,\color{green}{p}\,$ from the previous $\color{green}{y}$-values gives the new equation $\,\color{green}{y=f(x)-p}\,.$ This shifts the graph DOWN $\,\color{green}{p}\,$ units.

shifting a graph DOWN

A point $\,\color{purple}{(a,b)}\,$ on the graph of $\,\color{purple}{y=f(x)}\,$ moves to a point $\,\color{green}{(a,b-p)}\,$ on the graph of $\,\color{green}{y=f(x)-p}\,.$

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

Ideas Regarding Horizontal Translations (Moving Left/Right)

Summary of Horizontal Translations

Let $\,p\,$ be a positive number.

Start with the equation $\,\color{purple}{y=f(x)}\,.$ Replace every $\,\color{green}{x}\,$ by $\,\color{green}{x+p}\,$ to give the new equation $\,\color{green}{y=f(x+p)}\,.$ This shifts the graph LEFT $\,\color{green}{p}\,$ units.

shifting a graph left

A point $\,\color{purple}{(a,b)}\,$ on the graph of $\,\color{purple}{y=f(x)}\,$ moves to a point $\,\color{green}{(a-p,b)}\,$ on the graph of $\,\color{green}{y=f(x+p)}\,.$

Additionally:

Start with the equation $\,\color{purple}{y=f(x)}\,.$ Replace every $\,\color{green}{x}\,$ by $\,\color{green}{x-p}\,$ to give the new equation $\,\color{green}{y=f(x-p)}\,.$ This shifts the graph RIGHT $\,\color{green}{p}\,$ units.

shifting a graph to the right

A point $\,\color{purple}{(a,b)}\,$ on the graph of $\,\color{purple}{y=f(x)}\,$ moves to a point $\,\color{green}{(a+p,b)}\,$ on the graph of $\,\color{green}{y=f(x-p)}\,.$

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.)