Grade 11 · MCR3U · Functions

MCR3U: Transformations of Functions

Every transformation question in this unit is one form: $y = af[k(x - p)] + q$ . Learn what each letter does to the graph of $y = f(x)$ and you can graph, describe or reverse any of them.

How It Works

Outside the brackets is vertical. $a$ and $q$ act on the outputs, and they do what they look like. $a$ stretches vertically by a factor of $a$ , and a negative $a$ also reflects in the $x$ -axis. $q$ shifts up by $q$ .

Inside is horizontal, and backwards. $k$ and $p$ act on the inputs, and they do the opposite of what they look like. $k$ stretches horizontally by a factor of $\frac{1}{k}$ , so $k = 2$ squeezes the graph to half its width. A negative $k$ reflects in the $y$ -axis. $(x - p)$ shifts right by $p$ , so $f(x + 3)$ moves left $3$ .

**Factor $k$ out first.** $f(2x + 6)$ is not a shift of $6$ . Rewrite it as $f[2(x + 3)]$ : a horizontal compression by $\frac{1}{2}$ , then a shift left $3$ . Skipping this step is the most common mistake in the unit.

Order. Do reflections and stretches before translations. Stretching after shifting would stretch the shift too.

The mapping rule does all four at once: $(x, y) \to \left(\frac{x}{k} + p, ; ay + q\right)$ . Take a few key points of the parent function, push each through the rule, and join the images. For $y = \sqrt{x}$ use $(0, 0)$ , $(1, 1)$ and $(4, 2)$ ; for $y = x^2$ use the vertex and the points one and two units either side.

Domain and range follow the same rule. Push the endpoints of the domain through the $x$ part and the endpoints of the range through the $y$ part. A reflection swaps which end is the lower bound, so write the interval from smallest to largest.

Invariant points stay put. A vertical stretch leaves points on the $x$ -axis where they are; any translation moves every point.

5 Worked Examples

Easy first, test level last. Try each one on paper, then open the solution.

Easy
Describe how $y = f(x - 3) + 5$ transforms $y = f(x)$ , and find the image of the point $(2, 7)$ .
Show solution
1. Inside, $(x - 3)$ : shift right $3$ . Outside, $+5$ : shift up $5$ .
2. Mapping rule: $(x, y) \to (x + 3, y + 5)$ , so $(2, 7) \to (5, 12)$ .
Answer: Right $3$ , up $5$ ; $(2, 7) \to (5, 12)$ .
Easy
Describe $y = -2f(x)$ and find the image of $(4, 3)$ .
Show solution
1. $a = -2$ : a vertical stretch by a factor of $2$ and a reflection in the $x$ -axis. Inputs do not change.
2. Mapping rule: $(x, y) \to (x, -2y)$ , so $(4, 3) \to (4, -6)$ .
Answer: Vertical stretch by $2$ , reflection in the $x$ -axis; $(4, 3) \to (4, -6)$ .
Medium
Describe $y = f(2x + 6)$ and find the images of $(4, 1)$ and $(-2, 5)$ .
Show solution
1. Factor $k$ out: $f(2x + 6) = f[2(x + 3)]$ , so $k = 2$ and $p = -3$ .
2. Horizontal compression by a factor of $\frac{1}{2}$ , then a shift left $3$ .
3. Mapping rule: $(x, y) \to \left(\frac{x}{2} - 3, ; y\right)$ . So $(4, 1) \to (-1, 1)$ and $(-2, 5) \to (-4, 5)$ .
Answer: Compressed horizontally by $\frac{1}{2}$ , left $3$ ; images $(-1, 1)$ and $(-4, 5)$ .
Medium
For $f(x) = \sqrt{x}$ , graph $y = 3f[-(x - 1)] - 2$ from key points, and state its equation, domain and range.
Show solution
1. $a = 3$ , $k = -1$ , $p = 1$ , $q = -2$ : reflect in the $y$ -axis, stretch vertically by $3$ , shift right $1$ and down $2$ .
2. Mapping rule: $(x, y) \to (-x + 1, ; 3y - 2)$ .
3. Key points: $(0, 0) \to (1, -2)$ , $(1, 1) \to (0, 1)$ , $(4, 2) \to (-3, 4)$ .
4. The graph starts at $(1, -2)$ and runs left and up.
Answer: $y = 3\sqrt{-(x - 1)} - 2$ , through $(1, -2)$ , $(0, 1)$ , $(-3, 4)$ ; domain $x \le 1$ , range $y \ge -2$ .
Test level
$y = f(x)$ has domain $-2 \le x \le 4$ and range $-1 \le y \le 3$ , and passes through $(4, 3)$ . Find the domain and range of $y = -2f\left[\frac{1}{2}(x - 1)\right] + 3$ and the image of $(4, 3)$ .
Show solution
1. $a = -2$ , $k = \frac{1}{2}$ , $p = 1$ , $q = 3$ , so the mapping rule is $(x, y) \to (2x + 1, ; -2y + 3)$ .
2. Domain: $2(-2) + 1 = -3$ and $2(4) + 1 = 9$ , so $-3 \le x \le 9$ .
3. Range: $-2(-1) + 3 = 5$ and $-2(3) + 3 = -3$ . The reflection flips the order, so $-3 \le y \le 5$ .
4. Image: $(4, 3) \to (2(4) + 1, ; -2(3) + 3) = (9, -3)$ .
Answer: Domain $-3 \le x \le 9$ , range $-3 \le y \le 5$ ; $(4, 3) \to (9, -3)$ .

Learn It on Martian Lab

Martian Lab teaches this in the Transformations lesson of its MCR3U Functions & Transformations unit. Watch it, then take the Functions & Transformations practice tests to see if it stuck.

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Free Functions & Transformations practice test, with solutions

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