There are things that you can DO to an equation of the form
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$\,y=f(x)\,$
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
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$\,x\,$ 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:
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$\,y=(x3)^2+5\,$

Start with the graph of
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$\,y=x^2\,$.
(This is the ‘basic model’.)
 Add $\,5\,$ to the previous $\,y\,$values, giving the new equation
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$\,y=x^2+5\,$.
This moves the graph UP $\,5\,$ units.
 Replace every $\,x\,$ by
$\,x3\,$, giving the new equation
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$\,y=(x3)^2+5\,$.
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
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$\,x\,$ 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
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$\,\bigl(x,
\overset{\text{yvalue}}{\overbrace{
f(x)}}
\bigr) \,$.
Here, $\,x\,$ is the input, and $\,f(x)\,$ is the corresponding output.

The equation
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$\,y=f(x)\,$ 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
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$\,\bigl(x,
\overset{\text{yvalue}}{\overbrace{
f(x)}}
\bigr) \,$.

Compare the previous two ideas!
To ‘graph the function $\,f\,$’ means to show all points of the form
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$\,\bigl(x,f(x)\bigr)\,$.
To ‘graph the equation $\,y=f(x)\,$’ means to show all points of the form
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$\,\bigl(x,f(x)\bigr)\,$.
These two requests mean exactly the same thing!
IDEAS REGARDING VERTICAL TRANSLATIONS (MOVING UP/DOWN)
 Points on the graph of
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$\,y=f(x)\,$ are of the form $\,\bigl(x,f(x)\bigr)\,$.
Points on the graph of
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$\,y=f(x)+3\,$ 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
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$\,y=f(x)\,$
and asked about the graph of
$\,y=f(x)+3\,$:
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$$
\begin{align}
\text{original equation:} &\quad y=f(x)\cr\cr
\text{new equation:} &\quad y=f(x) + 3
\end{align}
$$
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$$
\begin{gather}
\text{interpretation of new equation:}\cr\cr
\overset{\text{the new yvalues}}{\overbrace{
\strut\ \ y\ \ }}
\overset{\text{are}}{\overbrace{
\strut\ \ =\ \ }}
\overset{\quad\text{the previous yvalues}\quad}{\overbrace{
\strut f(x)}}
\overset{\qquad\text{with 3 added to them}\quad}{\overbrace{
\strut\ \ + 3\ \ }}
\end{gather}
$$

Summary of vertical translations:
Let $\,p\,$ be a positive number.
Start with the equation
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$\,y=f(x)\,$.
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
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$\,y=f(x)\,$.
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 $\,(a,bp)\,$ 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
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$\,y=f(x)\,$ 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 $\,\color{red}{\bigl(x+3\,,\,f(x+3)\bigr)}\,$
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 counterintuitive—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 $\,y=f(x+3)\,$:
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$$
\begin{align}
\text{original equation:} &\quad y=f(x)\cr\cr
\text{new equation:} &\quad y=f(x+3)
\end{align}
$$
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$$
\begin{gather}
\text{interpretation of new equation:}\cr\cr
y = f(
\overset{\text{replace x by x+3}}{\overbrace{
x+3}}
)
\end{gather}
$$
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
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$\,y=f(x)\,$.
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 $\,(ap,b)\,$ on the graph of
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$\,y=f(x+p)\,$.
Additionally:
Start with the equation
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$\,y=f(x)\,$.
Replace every $\,x\,$ by $\,xp\,$ to
give the new equation
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$\,y=f(xp)\,$.
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
$\,y=f(xp)\,$.
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
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$\,y\,$
(that is, transformations that change the $\,y$values of the points),
we say:
DO THIS to the previous $\,y$value.
For transformations involving
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$\,x\,$
(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
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$\,y=f(x)\,$
to
$\,y = f(x) \pm c\,$
horizontal translations:
going from
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$\,y = f(x)\,$
to
$\,y = f(x\pm c)\,$
Make sure you see the difference between (say)
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$\,y = f(x) + 3\,$
and
$\,y = f(x+3)\,$!
In the case of
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$\,y = f(x) + 3\,$, 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
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$\,y = f(x + 3)\,$, 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 counterintuitive.
You must replace every $\,x\,$ by $\,x2\,$.
The new equation is:
$\,y = f(x2)\,$
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 counterintuitive.
Replacing every $\,x\,$ by $\,x+5\,$ in an equation causes the graph to shift $\,5\,$ units to the LEFT.
Thus, the new point is $\,(a5,b)\,$.