Shifting Graphs Up/Down/Left/Right


$y = f(x)$ $y = f(x) + 2\,$ up $2$	$y = f(x)$ $y = f(x) - 2\,$ down $2$	$y = f(x)$ $y = f(x+2)\,$ left $2$	$y = f(x)$ $y = f(x-2)\,$ right $2$
movements up and down change the $y$-values of points; transformations that affect the $y$-values are intuitive (e.g., to move up $\,2\,$, you add $\,2\,$ to the previous $\,y\,$-value)		movements left and right change the $x$-values of points; transformations that affect the $x$-values are counter-intuitive (e.g., 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)

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 $\,\color{purple}{y=f(x)}\,$.
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.
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.
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)

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 $\,\color{red}{P\bigl(x+3\,,\,f(x+3)\bigr)}\,$ 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.

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

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 $\,\color{purple}{y=f(x)}\,$.
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.
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.
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 $\,(a-5,b)\,$.

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

When you're done practicing, move on to:
horizontal and vertical stretching/shrinking

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

SHIFTING GRAPHS UP/DOWN/LEFT/RIGHT