Each single transformation was covered in Year 1 on Transformations of Graphs. Year 2 asks you to combine them.
| Equation | Transformation of $y = \mathrm{f}(x)$ | Point $(p, q)$ moves to |
|---|---|---|
| $y = \mathrm{f}(x) + d$ | Translation by $\begin{pmatrix} 0 \\ d \end{pmatrix}$ | $(p,\; q + d)$ |
| $y = \mathrm{f}(x + c)$ | Translation by $\begin{pmatrix} -c \\ 0 \end{pmatrix}$ | $(p - c,\; q)$ |
| $y = a\,\mathrm{f}(x)$ | Vertical stretch, scale factor $a$ | $(p,\; aq)$ |
| $y = \mathrm{f}(bx)$ | Horizontal stretch, scale factor $\dfrac{1}{b}$ | $\left(\dfrac{p}{b},\; q\right)$ |
| $y = -\mathrm{f}(x)$ | Reflection in the $x$-axis | $(p,\; -q)$ |
| $y = \mathrm{f}(-x)$ | Reflection in the $y$-axis | $(-p,\; q)$ |
- Horizontal changes (inside the bracket): translate by $-c$ first, then stretch by $\tfrac{1}{b}$. (Or write $b\left(x + \tfrac{c}{b}\right)$ and stretch first, then translate by $-\tfrac{c}{b}$.)
- Vertical changes (outside): stretch by $a$ (reflect if $a < 0$) first, then translate by $d$.
- Horizontal and vertical changes do not interfere with each other, so these two groups can be done in either order.
Describe how to obtain $y = \mathrm{f}(2x - 6)$ from $y = \mathrm{f}(x)$, and find the image of $(4, 1)$.
The curve $y = \mathrm{f}(x)$ has a maximum at $(2, 5)$ and crosses the $y$-axis at $(0, 3)$. Find the corresponding points on $y = 3\mathrm{f}(x + 1) - 2$.
Describe a sequence of transformations mapping $y = x^2$ onto $y = -2(x - 1)^2 + 4$.
Describe the transformations from $y = \sin x$ to $y = 3\sin(2x + 60^\circ)$, and find the first maximum with $x > 0$.
| Graph | How to draw it | Symmetry |
|---|---|---|
| $y = |\mathrm{f}(x)|$ | Reflect the parts below the $x$-axis in the $x$-axis | Never below the $x$-axis |
| $y = \mathrm{f}(|x|)$ | Keep the part for $x \geqslant 0$, delete the part for $x < 0$, then reflect the right-hand part in the $y$-axis | Symmetric in the $y$-axis |
$\mathrm{f}(x) = x^2 - 4x + 3$. Sketch $y = \mathrm{f}(|x|)$ and $y = |\mathrm{f}(x)|$, giving turning points and intercepts.
Sketch $y = 2|x - 1| - 4$, giving the vertex and the intercepts.
Outside
Acts on $y$, does what it says.
Inside
Acts on $x$, does the opposite.
Point rule
$(p,q) \mapsto \left(\dfrac{p-c}{b},\; aq+d\right)$.
Vertical order
Stretch, then translate.
Horizontal order
Translate by $-c$, then stretch by $\tfrac1b$.
Safest method
New $x$: solve $bx + c = p$.
$|\mathrm{f}(x)|$
Reflect below-axis parts upward.
$\mathrm{f}(|x|)$
Copy the right half onto the left.
Max or min?
Swaps if $a < 0$.
The point $(3, -2)$ lies on $y = \mathrm{f}(x)$. Find its image on (a) $y = \mathrm{f}(x - 2) + 5$ (b) $y = 3\mathrm{f}(2x)$ (c) $y = 1 - \mathrm{f}(x)$.
▶ Show solution
(a) $x - 2 = 3 \Rightarrow x = 5$; $y = -2 + 5 = 3$. Image $(5, 3)$.
(b) $2x = 3 \Rightarrow x = 1.5$; $y = 3(-2) = -6$. Image $(1.5, -6)$.
(c) $y = 1 - (-2) = 3$. Image $(3, 3)$.
Describe fully the two transformations that map $y = \mathrm{f}(x)$ onto $y = \mathrm{f}(3x) - 4$.
▶ Show solution
A stretch parallel to the $x$-axis with scale factor $\tfrac13$, and a translation by $\begin{pmatrix} 0 \\ -4 \end{pmatrix}$ (in either order, since one is horizontal and one vertical).
Describe a sequence of transformations that maps $y = x^2$ onto $y = 4 - (x + 2)^2$, and state the vertex.
▶ Show solution
Translate by $\begin{pmatrix} -2 \\ 0 \end{pmatrix}$: $y = (x + 2)^2$.
Reflect in the $x$-axis: $y = -(x + 2)^2$.
Translate by $\begin{pmatrix} 0 \\ 4 \end{pmatrix}$: $y = 4 - (x + 2)^2$.
Vertex $(-2, 4)$, a maximum.
$y = \mathrm{f}(x)$ has a maximum point at $(4, 6)$. State the coordinates of the corresponding turning point, and its nature, on (a) $y = 2\mathrm{f}(x - 1) + 3$ (b) $y = \mathrm{f}(2x) - 6$ (c) $y = -\mathrm{f}(x)$.
▶ Show solution
(a) $(5, 15)$, maximum.
(b) $(2, 0)$, maximum.
(c) $(4, -6)$, now a minimum because of the reflection.
The point $(6, -1)$ lies on $y = \mathrm{f}(x)$. Find its image on $y = \mathrm{f}(2x + 4)$, and describe the transformation in two different orders.
▶ Show solution
$2x + 4 = 6 \Rightarrow x = 1$. Image $(1, -1)$.
Order 1: translate by $\begin{pmatrix} -4 \\ 0 \end{pmatrix}$, then stretch horizontally by $\tfrac12$: $(6, -1) \to (2, -1) \to (1, -1)$ ✓.
Order 2: $\mathrm{f}\big(2(x + 2)\big)$: stretch by $\tfrac12$, then translate by $\begin{pmatrix} -2 \\ 0 \end{pmatrix}$: $(6, -1) \to (3, -1) \to (1, -1)$ ✓.
The graph of $y = \mathrm{f}(x)$ is translated by $\begin{pmatrix} 3 \\ 0 \end{pmatrix}$ and then stretched parallel to the $x$-axis with scale factor $2$. Write down the equation of the new graph. Is it the same as doing the stretch first?
▶ Show solution
After the translation: $y = \mathrm{f}(x - 3)$.
Stretching by $2$ replaces $x$ by $\tfrac{x}{2}$: $y = \mathrm{f}\left(\tfrac{x}{2} - 3\right)$.
Stretch first gives $y = \mathrm{f}\left(\tfrac{x}{2}\right)$, then translating gives $y = \mathrm{f}\left(\tfrac{x - 3}{2}\right) = \mathrm{f}\left(\tfrac{x}{2} - \tfrac32\right)$. These are different.
Sketch $y = 3|x + 1| - 6$, giving the vertex and the points where it crosses the axes.
▶ Show solution
Vertex $(-1, -6)$.
$x$-intercepts: $|x + 1| = 2 \Rightarrow x = 1$ or $x = -3$.
$y$-intercept: $3(1) - 6 = -3$, so $(0, -3)$.
A V shape with arms of gradient $\pm 3$.
$\mathrm{f}(x) = (x - 2)(x + 1)$. Sketch (a) $y = |\mathrm{f}(x)|$ and (b) $y = \mathrm{f}(|x|)$, giving the coordinates of turning points and intercepts.
▶ Show solution
$\mathrm{f}(x) = x^2 - x - 2 = \left(x - \tfrac12\right)^2 - \tfrac94$: roots $-1, 2$; minimum $\left(\tfrac12, -\tfrac94\right)$; $y$-intercept $-2$.
(a) Reflect the part between $-1$ and $2$: maximum $\left(\tfrac12, \tfrac94\right)$, touching the axis at $(-1, 0)$ and $(2, 0)$, $y$-intercept $(0, 2)$.
(b) Keep $x \geqslant 0$ (root $2$, minimum $\left(\tfrac12, -\tfrac94\right)$, $y$-intercept $-2$) and reflect in the $y$-axis.
Roots $x = \pm 2$; minima $\left(\pm\tfrac12, -\tfrac94\right)$; a sharp local maximum at $(0, -2)$. The root at $-1$ disappears.
For $0^\circ \leqslant x \leqslant 360^\circ$, sketch $y = 2\cos(x - 60^\circ) + 1$. State its maximum and minimum points, its $y$-intercept, and the values of $x$ where it crosses the $x$-axis.
▶ Show solution
From $y = \cos x$: translate by $\begin{pmatrix} 60^\circ \\ 0 \end{pmatrix}$, stretch vertically by $2$, translate by $\begin{pmatrix} 0 \\ 1 \end{pmatrix}$.
Maximum when $x - 60^\circ = 0^\circ$: $(60^\circ, 3)$. Minimum when $x - 60^\circ = 180^\circ$: $(240^\circ, -1)$.
$y$-intercept: $2\cos(-60^\circ) + 1 = 2 \times \tfrac12 + 1 = 2$, so $(0, 2)$.
$y = 0$: $\cos(x - 60^\circ) = -\tfrac12 \Rightarrow x - 60^\circ = 120^\circ$ or $240^\circ$, so $x = 180^\circ$ or $x = 300^\circ$.
$\mathrm{f}(x) = 2x^2 - 12x + 13$.
(a) Write $\mathrm{f}(x)$ in the form $a(x + p)^2 + q$. (b) Describe a sequence of transformations mapping $y = x^2$ onto $y = \mathrm{f}(x)$. (c) Find the exact roots of $\mathrm{f}(x) = 0$. (d) Sketch $y = |\mathrm{f}(x)|$ and state the number of solutions of $|\mathrm{f}(x)| = 5$. (e) State the number of solutions of $\mathrm{f}(|x|) = 13$.
▶ Show solution
(a) $2(x^2 - 6x) + 13 = 2\big((x - 3)^2 - 9\big) + 13 = 2(x - 3)^2 - 5$.
(b) Translate by $\begin{pmatrix} 3 \\ 0 \end{pmatrix}$, stretch vertically by scale factor $2$, then translate by $\begin{pmatrix} 0 \\ -5 \end{pmatrix}$.
(c) $(x - 3)^2 = \tfrac52 \Rightarrow x = 3 \pm \sqrt{\tfrac52} = 3 \pm \dfrac{\sqrt{10}}{2}$.
(d) The minimum $(3, -5)$ reflects to a local maximum $(3, 5)$, between the roots in (c); $y$-intercept $13$.
The line $y = 5$ touches the local maximum and crosses each outer arm once: three solutions.
(e) For $x \geqslant 0$: $2x^2 - 12x + 13 = 13 \Rightarrow 2x(x - 6) = 0 \Rightarrow x = 0$ or $6$.
Reflecting in the $y$-axis adds $x = -6$; $x = 0$ is its own mirror image. So three solutions: $x = -6, 0, 6$.