Graph Transformations

In short: A graph transformation moves, flips or stretches a graph. $f (x) + c$ moves it up by $c$ , $f (x - b)$ moves it right by $b$ , $- f (x)$ flips it in the x-axis, and $k f (x)$ stretches it vertically. Changes inside the bracket do the opposite of what they look like.

Starting point: Years 9 and 10

Start with $y = x^{2}$ , vertex at the origin.

$y = x^{2} + 3$ moves it up 3.
$y = x^{2} - 3$ moves it down 3.
$y = (x - 2)^{2}$ moves it right 2.
$y = (x + 2)^{2}$ moves it left 2.
$y = - x^{2}$ flips it upside down.
$y = 3 x^{2}$ makes it narrower (steeper). $y = \frac{1}{3} x^{2}$ makes it wider.

Right 2 and up 1: the vertex moves from (0, 0) to (2, 1).

Transformations

Translations

f(x) + c: up c
f(x - b): right b

Reflections

-f(x): in the x-axis
f(-x): in the y-axis

Dilations

k f(x): vertical stretch by k
f(ax): horizontal by 1/a

Combined

y = k f(a(x + b)) + c
Inside first, then outside

Year 12Combining transformations

The general form is

$y = k f (a (x + b)) + c$

Work from the inside out:

Horizontal dilation by $\frac{1}{a}$ (and reflection in the y-axis if $a < 0$ ).
Horizontal translation by $- b$ .
Vertical dilation by $k$ (and reflection in the x-axis if $k < 0$ ).
Vertical translation by $c$ .

Mapping a point. Under $y = 2 f (3 (x - 1)) + 5$ , the point $(6, 4)$ on $y = f (x)$ goes to $x = \frac{6}{3} + 1 = 3$ and $y = 2 (4) + 5 = 13$ . New point: $(3, 13)$ .

Watch the brackets. $f (2 x - 6) = f (2 (x - 3))$ is a shift right 3, not 6. Always factor out $a$ first.

Transformations also change the domain and range. If $f$ has range $[0, 2]$ , then $3 f (x) - 1$ has range $[- 1, 5]$ .

Working backwards. To find the original point from an image, undo the steps in reverse order. If $(5, 2)$ is on $y = g (3 (x - 1)) - 4$ , then the original $y$ is $2 + 4 = 6$ and the original $x$ is $3 (5 - 1) = 12$ , so the point was $(12, 6)$ .

Reflections with a shift. $f (2 - x) = f (- (x - 2))$ : reflect in the $y$ -axis, then move right 2.

From an expanded equation. Complete the square to see the transformations: $y = 2 x^{2} + 12 x + 13 = 2 (x + 3)^{2} - 5$ is $y = x^{2}$ stretched vertically by 2, moved left 3 and down 5.

Logs and exponentials can be rewritten. $ln (2 x - 2) = ln 2 + ln (x - 1)$ , so the same graph is $y = ln x$ moved right 1 and up $ln 2$ . And $e^{x + l n 3} = 3 e^{x}$ , so moving $y = e^{x}$ left $ln 3$ is the same as stretching it vertically by 3.

Tangents move too. A translation does not change gradients. If the tangent to $y = f (x)$ at $x = 1$ is $y = 2 x + 1$ , then $y = f (x - 3) + 4$ has a tangent of gradient 2 at $x = 4$ , through the moved point $(4, 7)$ : $y = 2 x - 1$ .

Quick summary

$f (x) + c$ moves up $c$ . $f (x - b)$ moves right $b$ .
$- f (x)$ flips in the x-axis. $f (- x)$ flips in the y-axis.
$k f (x)$ stretches vertically by $k$ . $f (a x)$ stretches horizontally by $\frac{1}{a}$ .
Changes inside the bracket act on $x$ and do the opposite of what they look like.
For $y = k f (a (x + b)) + c$ , factor out $a$ and work from the inside out.

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How is $y = x^{2} + 5$ related to $y = x^{2}$ ?
1. A moved up 5
2. B moved down 5
3. C moved right 5
4. D moved left 5
Show answer
A: moved up 5
Adding outside the function moves it up.
Where is the vertex of $y = (x - 3)^{2}$ ?
1. A $(- 3, 0)$
2. B $(3, 0)$
3. C $(0, 3)$
4. D $(0, - 3)$
Show answer
B: $(3, 0)$
The bracket is zero at $x = 3$ , so it moves right 3.
Where is the vertex of $y = (x + 2)^{2} - 7$ ?
1. A $(2, - 7)$
2. B $(2, 7)$
3. C $(- 2, - 7)$
4. D $(- 2, 7)$
Show answer
C: $(- 2, - 7)$
Left 2 and down 7.

Starting point: Years 9 and 10

Quick summary

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