Pure Mathematics 1

Mathematics9709

Pure Mathematics 1

Mathematics9709

The Four Function Transformations

A graph changes a known graph. A translation moves it, a reflection flips it in a line, and a stretch multiplies distances in one direction.

Outside f changes outputs, so it acts vertically as written. Inside f changes inputs, so it acts horizontally with the opposite sign or reciprocal factor.

The four transformations

“Parallel to an axis” names the direction in which point coordinates are multiplied. Check whether the change is outside or inside f. In a translation vector $\binom{p}{q}$ , the top number is horizontal and the bottom number is vertical; positive means right or up.

$\mathrm{f}(x)+a$ : translation by $\binom{0}{a}$ . Every point $(x,y)$ becomes $(x,y+a)$ .
$\mathrm{f}(x-a)$ : translation right by $a$ , vector $\binom{a}{0}$ . Therefore $\mathrm{f}(x+a)$ moves left by $a$ .
For $a>0$ , $a\mathrm{f}(x)$ : stretch by factor $a$ parallel to the $y$ -axis. It maps $(x,y)\to(x,ay)$ .
For $a>0$ , $\mathrm{f}(ax)$ : stretch by factor $\tfrac1a$ parallel to the $x$ -axis. It maps $(x,y)\to(\tfrac{x}{a},y)$ .

$-\mathrm{f}(x)$ reflects in the $x$ -axis; $\mathrm{f}(-x)$ reflects in the $y$ -axis. More generally, $\mathrm{f}(-ax)$ for $a>0$ also stretches horizontally by factor $\tfrac1a$ .
$a=0$ is not a stretch: $0\mathrm{f}(x)=0$ , while $\mathrm{f}(0x)=\mathrm{f}(0)$ when f is defined at zero.

Worked example

Track one point through each reflection

If $(2,5)$ lies on $y=\mathrm{f}(x)$ , then reflection in the $x$ -axis gives $(2,-5)$ , while reflection in the $y$ -axis gives $(-2,5)$ .

Answer

-\mathrm{f}(x)

(2,-5)

\mathrm{f}(-x)

(-2,5)

These rules depend on where the change is written, not on the graph's shape. They apply in the same way to polynomial, reciprocal, square-root and trigonometric graphs.

Drag it: watch a, b and c reshape the curve

Here f is $x^2$ . Change $a$ , $b$ and $c$ ; the transformed vertex is $(b,c)$ .

a (stretch)b (left/right)c (up/down)

y = (x + 0)2 + 0·vertex (0, 0)·the starting curve, unchanged

Drag a (vertical stretch), b (left and right, the inside one) and c (up and down): the curve changes as you drag, and the vertex moves to (b, c).

Set $b=2$ : the bracket is $x-2$ and the vertex moves right to $x=2$ . This is the inside-sign rule.

Your turnIndependent transformation check

State the single translation that maps $y=\mathrm{f}(x)$ onto $y=\mathrm{f}(x-4)+3$ .

Show worked answer

The $x-4$ is inside, so the graph moves right 4. The $+3$ is outside, so it moves up 3.

Answer

\text{translation by }\binom{4}{3}

PreviousThe graph of a function and its inverse NextDescribing Function Transformations

The Four Function Transformations

A graph changes a known graph. A translation moves it, a reflection flips it in a line, and a stretch multiplies distances in one direction.

Outside f changes outputs, so it acts vertically as written. Inside f changes inputs, so it acts horizontally with the opposite sign or reciprocal factor.

The four transformations

$\mathrm{f}(x)+a$ : translation by $\binom{0}{a}$ . Every point $(x,y)$ becomes $(x,y+a)$ .
$\mathrm{f}(x-a)$ : translation right by $a$ , vector $\binom{a}{0}$ . Therefore $\mathrm{f}(x+a)$ moves left by $a$ .
For $a>0$ , $a\mathrm{f}(x)$ : stretch by factor $a$ parallel to the $y$ -axis. It maps $(x,y)\to(x,ay)$ .
For $a>0$ , $\mathrm{f}(ax)$ : stretch by factor $\tfrac1a$ parallel to the $x$ -axis. It maps $(x,y)\to(\tfrac{x}{a},y)$ .

$-\mathrm{f}(x)$ reflects in the $x$ -axis; $\mathrm{f}(-x)$ reflects in the $y$ -axis. More generally, $\mathrm{f}(-ax)$ for $a>0$ also stretches horizontally by factor $\tfrac1a$ .
$a=0$ is not a stretch: $0\mathrm{f}(x)=0$ , while $\mathrm{f}(0x)=\mathrm{f}(0)$ when f is defined at zero.

Worked example

Track one point through each reflection

If $(2,5)$ lies on $y=\mathrm{f}(x)$ , then reflection in the $x$ -axis gives $(2,-5)$ , while reflection in the $y$ -axis gives $(-2,5)$ .

Answer

-\mathrm{f}(x)

(2,-5)

\mathrm{f}(-x)

(-2,5)

These rules depend on where the change is written, not on the graph's shape. They apply in the same way to polynomial, reciprocal, square-root and trigonometric graphs.

Drag it: watch a, b and c reshape the curve

Here f is $x^2$ . Change $a$ , $b$ and $c$ ; the transformed vertex is $(b,c)$ .

a (stretch)b (left/right)c (up/down)

y = (x + 0)2 + 0·vertex (0, 0)·the starting curve, unchanged

Drag a (vertical stretch), b (left and right, the inside one) and c (up and down): the curve changes as you drag, and the vertex moves to (b, c).

Set $b=2$ : the bracket is $x-2$ and the vertex moves right to $x=2$ . This is the inside-sign rule.

Your turnIndependent transformation check

State the single translation that maps $y=\mathrm{f}(x)$ onto $y=\mathrm{f}(x-4)+3$ .

Show worked answer

The $x-4$ is inside, so the graph moves right 4. The $+3$ is outside, so it moves up 3.

Answer

\text{translation by }\binom{4}{3}

PreviousThe graph of a function and its inverse NextDescribing Function Transformations