Transforming Function Graphs

A function $f$ has domain $[2, 6]$ and range $[0, 9]$. Before reading on, predict the domain and range of $y = f(x) + 4$, and then of $y = f(x-4)$. One of your two answers should have a changed domain and the other a changed range.

The four transformations act on the graph, not on the formula. A change written outside the function moves the outputs, so it moves the range. A change written inside moves the inputs, so it moves the domain, and appears reversed.

Every transformation in this lesson is defined by what it does to the outputs or to the inputs of $f$, and neither depends on how $f$ is calculated.

Take $y = f(x) + 2$. Whatever height the original graph had at a given $x$, the new graph has two more. That statement is complete without knowing what $f$ is, so the whole graph can be redrawn from a picture alone.

The diagram above does exactly that: an unlabelled curve and its image under a shift of two right and two up. No formula was needed at either stage.

This matters more than it might appear. It means:

a question can give you a sketch of an unknown $f$ and ask for a transformed sketch;
a question can give you the domain and range of an unknown $f$ and ask for the transformed ones;
and every rule you learn here applies to functions you have not met yet, including those in senior courses.

The rules are about graphs, not about algebra. That is why they are worth learning once, properly.

Three of the four transformations are applied to the output after $f$ has finished.

$y = f(x) + c$. Every height increases by $c$, so the graph slides up by $c$ (down if $c$ is negative). The domain is untouched, since the same inputs are still being fed in. The range shifts by $c$.

$y = -f(x)$. Every height reverses sign, so the graph flips across the horizontal axis. The domain is untouched. The range reflects about zero, and its inequalities reverse: $[1,7]$ becomes $[-7,-1]$.

$y = kf(x)$. Every height is multiplied by $k$, so the graph stretches away from the horizontal axis if $|k| > 1$ and squashes towards it if $|k| < 1$. A negative $k$ reflects as well. The domain is untouched. The range is multiplied by $k$.

The pattern is uniform: whatever you do to the output, do the same to the range. If $f$ has range $[1,7]$, then $3f(x)$ has range $[3,21]$, $f(x)-5$ has range $[-4,2]$, and $-f(x)$ has range $[-7,-1]$.

One transformation is applied to the input before $f$ ever sees it, and it behaves differently in two ways.

$y = f(x-a)$ translates the graph $a$ units to the right.

It looks reversed, and here is why. Call the new function $g$, so $g(x) = f(x-a)$. For $g$ to do at some input whatever $f$ did at input $t$, the bracket must equal $t$: that is $x - a = t$, so $x = t + a$. Every feature of $f$ therefore reappears $a$ further along.

The consequence for the domain follows immediately. If $f$ accepts inputs in $[2,6]$, then $g$ accepts $x$ exactly when $x - a$ lies in $[2,6]$, that is when $x$ lies in $[2+a, 6+a]$. The domain shifts by $a$, in the same direction as the graph.

The range is untouched, because $g$ produces exactly the same outputs as $f$; it merely produces each one at a different input.

So the two halves of the rule pull apart cleanly:

$$f(x-a): \quad \text{domain} + a, \quad \text{range unchanged}$$

$$f(x)+c: \quad \text{domain unchanged}, \quad \text{range} + c$$

Combining this with Lesson 6 makes most sketching questions routine. Read the transformations off the equation, then apply them to the base shape's domain and range.

$y = 2\sqrt{x-1} + 3$. Base $\sqrt{x}$, domain $[0,\infty)$, range $[0,\infty)$. The $-1$ is inside, so the domain becomes $[1,\infty)$. Outside, the range is multiplied by $2$, still $[0,\infty)$, then raised by $3$ to $[3,\infty)$. The graph is the usual root shape, starting at $(1,3)$ and rising, twice as steeply as the base.

$y = -(x+2)^2 + 5$. Base $x^2$, domain all reals, range $[0,\infty)$. The $+2$ inside shifts left by $2$, and the domain is still all real numbers. Outside, the minus reflects the range to $(-\infty,0]$, then the $+5$ raises it to $(-\infty,5]$. The graph is a downward parabola with vertex $(-2,5)$.

$y = \dfrac{1}{x-4} + 1$. Base $\dfrac{1}{x}$, domain $x \neq 0$, range $y \neq 0$. Inside, the domain becomes $x \neq 4$. Outside, the range becomes $y \neq 1$. The asymptotes are therefore $x = 4$ and $y = 1$, and each corresponds to one of those exclusions.

In every case the base shape survives unchanged; only its position and scale differ.

Five items on the four transformations. Work each one, then reveal the answer.

1. 1 $f$ has domain $[0,4]$. State the domain of $y = f(x) + 9$.

   Show answer
   The change is outside, so the inputs are untouched.$[0,4]$

2. 2 $f$ has domain $[0,4]$. State the domain of $y = f(x-3)$.

   Show answer
   The change is inside, so the domain shifts right by $3$.$[3,7]$

3. 3 $f$ has range $[1,5]$. State the range of $y = -f(x)$.

   Show answer
   Reflect both endpoints and put them in order.$[-5,-1]$

4. 4 $f$ has range $[2,6]$. State the range of $y = 4f(x)$.

   Show answer
   Multiply both endpoints by $4$.$[8,24]$

5. 5 Describe the change from $y = f(x)$ to $y = f(x+7)$.

   Show answer
   Inside the brackets, so the effect is reversed.Translate $7$ units left