$f(x)$ — the original function (the starting graph)
$f(x) + a$ — add $a$ to every output (y-value) of $f$
$f(x + a)$ — replace every $x$ input with $x + a$ before applying $f$
$-f(x)$ — negate every output (flip all y-values)
$f(-x)$ — replace every $x$ input with $-x$ before applying $f$
| Transformation | Notation | Type | Direction |
|---|---|---|---|
| Shift up by $a$ | $f(x) + a$ | Translation | Vertical (outside) |
| Shift right by $a$ | $f(x - a)$ | Translation | Horizontal (inside) |
| Reflection in x-axis | $-f(x)$ | Reflection | Vertical (outside) |
| Reflection in y-axis | $f(-x)$ | Reflection | Horizontal (inside) |
Shifts the graph up by $a$ units if $a > 0$, or down by $|a|$ units if $a < 0$.
Every point $(x,\ y)$ on $f(x)$ moves to $(x,\ y + a)$.
$f(x)+3$ (red) shifts up 3; $f(x)-3$ (blue) shifts down 3. Original $f(x)=x^2$ shown dashed.
How to apply a vertical translation
- Identify the value of $a$ (positive = up, negative = down).
- Add $a$ to the y-coordinate of every key point (roots, turning point, intercepts).
- x-coordinates of all points stay the same — only y-values change.
- Draw the same shape curve, shifted vertically.
Shifts the graph left by $a$ units if $a > 0$, or right by $|a|$ units if $a < 0$.
Every point $(x,\ y)$ on $f(x)$ moves to $(x - a,\ y)$.
$f(x + 3)$ moves the graph LEFT by 3, NOT right.
$f(x - 3)$ moves the graph RIGHT by 3, NOT left.
Think of it this way: $f(x+3)$ reaches the same $y$-value when $x$ is 3 less than before — so the graph moves left.
$f(x-2)$ (red) shifts right 2; $f(x+2)$ (blue) shifts left 2. Note: vertices marked with dots.
How to apply a horizontal translation
- Identify the value of $a$ in $f(x + a)$.
- Every point $(x, y)$ moves to $(x - a,\ y)$ — subtract $a$ from every x-coordinate.
- y-coordinates of all points stay the same — only x-values change.
- Draw the same shape curve, shifted horizontally.
The translation vector shows the shift: $-a$ in the x-direction, $+b$ in the y-direction.
Every y-value is negated. Points on the x-axis stay fixed.
Each point $(x,\ y)$ maps to $(x,\ -y)$.
$y = f(x) = x^2 - 2x$ (original)
$y = -f(x) = -(x^2 - 2x)$ (reflected)
$-f(x)$ flips the curve in the x-axis: roots stay fixed, the minimum $(1,-1)$ becomes a maximum $(1, 1)$, and U-shape becomes ∩-shape.
What changes: Maxima become minima, minima become maxima. U-shapes become ∩-shapes and vice versa.
Every x-value is negated. Points on the y-axis stay fixed.
Each point $(x,\ y)$ maps to $(-x,\ y)$.
$y = f(x) = x^2 - 2x$ (original)
$y = f(-x) = x^2 + 2x$ (reflected)
$f(-x)$ reflects in the y-axis: the curve is mirrored left↔right. Root at $x=2$ moves to $x=-2$; vertex at $(1,-1)$ moves to $(-1,-1)$.
What changes: Every root at $x = r$ moves to $x = -r$. The shape is mirrored left to right.
It's very efficient to track how transformations move specific coordinates. In the exam you will often be given a point on $f(x)$ and asked for its image under a transformation.
| Transformation | Point $(x,\ y)$ maps to… | x changes? | y changes? |
|---|---|---|---|
| $f(x) + a$ | $(x,\ y + a)$ | No | Yes, add $a$ |
| $f(x - a)$ | $(x + a,\ y)$ | Yes, add $a$ | No |
| $f(x + a)$ | $(x - a,\ y)$ | Yes, subtract $a$ | No |
| $f(x+a)+b$ | $(x - a,\ y + b)$ | Yes, subtract $a$ | Yes, add $b$ |
| $-f(x)$ | $(x,\ -y)$ | No | Yes, negate |
| $f(-x)$ | $(-x,\ y)$ | Yes, negate | No |
Find the image of $(4, 5)$ under each transformation:
$f(x) + a$
Vertical shift
$a>0$: up by $a$
$a<0$: down by $|a|$
$(x,y) \to (x, y+a)$
$f(x - a)$
Horizontal shift right by $a$
$a>0$: right
$a<0$: left
$(x,y) \to (x+a, y)$
$f(x + a)$
Horizontal shift LEFT by $a$
⚠️ Opposite to sign!
$(x,y) \to (x-a, y)$
$f(x+a)+b$
Combined translation
Vector $\begin{pmatrix}-a\\b\end{pmatrix}$
$(x,y) \to (x-a, y+b)$
$-f(x)$
Reflect in x-axis
y-values negated
$(x,y) \to (x, -y)$
Roots stay fixed
$f(-x)$
Reflect in y-axis
x-values negated
$(x,y) \to (-x, y)$
y-intercept stays fixed
Describe the transformation that maps $y = f(x)$ onto each of the following:
(a) $y = f(x) + 5$ (b) $y = f(x - 4)$ (c) $y = -f(x)$ (d) $y = f(x + 2) - 1$ (e) $y = f(-x)$
▶ Show solution
(a) $y = f(x) + 5$: Translation up by 5. Vector $\begin{pmatrix}0\\5\end{pmatrix}$.
(b) $y = f(x-4)$: Translation right by 4. Vector $\begin{pmatrix}4\\0\end{pmatrix}$.
(c) $y = -f(x)$: Reflection in the x-axis.
(d) $y = f(x+2)-1$: Translation left 2 and down 1. Vector $\begin{pmatrix}-2\\-1\end{pmatrix}$.
(e) $y = f(-x)$: Reflection in the y-axis.
The graph of $y = x^2$ is transformed. Write the equation of the new graph after each transformation:
(a) Translate up by 6 (b) Translate right by 5 (c) Translate left 2, down 3 (d) Translate by vector $\begin{pmatrix}4\\-1\end{pmatrix}$
▶ Show solution
(a) $y = x^2 + 6$
(b) $y = (x-5)^2$
(c) Translate left 2 → $f(x+2)$; down 3 → $-3$. So: $y = (x+2)^2 - 3$
(d) Vector $\begin{pmatrix}4\\-1\end{pmatrix}$ means right 4, down 1: $y = (x-4)^2 - 1$
The point $(3, -2)$ lies on the graph of $y = f(x)$. Find the coordinates of the image of this point on the graph of:
(a) $y = f(x) - 5$ (b) $y = f(x + 4)$ (c) $y = -f(x)$ (d) $y = f(-x)$ (e) $y = f(x-1) + 3$
▶ Show solution
(a) $f(x)-5$: $y$ decreases by 5 → $(3,\ -7)$
(b) $f(x+4)$: $x$ decreases by 4 → $(-1,\ -2)$
(c) $-f(x)$: $y$ negated → $(3,\ 2)$
(d) $f(-x)$: $x$ negated → $(-3,\ -2)$
(e) $f(x-1)+3$: $x$ increases by 1, $y$ increases by 3 → $(4,\ 1)$
The graph of $y = \sin x$ is translated by vector $\begin{pmatrix}90\\0\end{pmatrix}$. Write the equation of the resulting graph and describe how it relates to $y = \cos x$.
▶ Show solution
A translation right by 90° is achieved by replacing $x$ with $x - 90°$:
$$y = \sin(x - 90°)$$
Since $\sin(x - 90°) = -\cos x$, the resulting graph is $y = -\cos x$.
Alternatively, $y = \cos x$ is $y = \sin x$ translated left by 90°: $\sin(x + 90°) = \cos x$.
The graph of $y = x^2 - 6x + 5$ has roots at $x = 1$ and $x = 5$, and a minimum at $(3, -4)$.
(a) Write the equation of its reflection in the x-axis. State the new turning point and roots.
(b) Write the equation of its reflection in the y-axis. State the new roots.
▶ Show solution
(a) Reflection in x-axis → $y = -(x^2 - 6x + 5) = -x^2 + 6x - 5$
Turning point: $(3, -4) \to (3, 4)$ — becomes a maximum at $(3, 4)$.
Roots: unchanged at $x = 1$ and $x = 5$ (y-values were already 0).
(b) Reflection in y-axis → replace $x$ with $-x$: $y = (-x)^2 - 6(-x) + 5 = x^2 + 6x + 5$
New roots: solve $x^2 + 6x + 5 = 0 \Rightarrow (x+1)(x+5) = 0 \Rightarrow x = -1$ and $x = -5$.
(These are the negatives of the original roots $1$ and $5$, as expected.)
Show that $y = x^2 + 4x + 7$ can be written as $y = (x+2)^2 + 3$. Hence describe the transformation that maps $y = x^2$ onto $y = x^2 + 4x + 7$.
▶ Show solution
Complete the square: $y = (x+2)^2 - 4 + 7 = (x+2)^2 + 3$ ✓
This is in the form $f(x+2)+3$ where $f(x) = x^2$.
So the transformation is: translation left 2, up 3.
Translation vector: $\begin{pmatrix}-2\\3\end{pmatrix}$.
The vertex moves from $(0,0)$ to $(-2, 3)$.
Graph A has the equation $y = f(x)$ and passes through the points $(-2, 0)$, $(0, -4)$, $(2, 0)$ and has minimum at $(0, -4)$.
Graph B passes through $(-2, 4)$, $(0, 0)$, $(2, 4)$ with minimum at $(0, 0)$.
Describe the transformation that maps Graph A to Graph B.
▶ Show solution
Compare key points. Minimum of A is $(0,-4)$; minimum of B is $(0,0)$.
The x-coordinate of the minimum hasn't changed, but y went from $-4$ to $0$ — an increase of $4$.
Check another point: $(-2, 0)$ on A maps to $(-2, 4)$ on B — y increased by 4. ✓
Transformation: $y = f(x) + 4$ — a vertical translation up by 4.
The graph of $y = 3^x$ passes through $(0, 1)$, $(1, 3)$ and $(2, 9)$.
(a) Write the equation of its reflection in the x-axis. Find three points on this reflected graph.
(b) Write the equation of its reflection in the y-axis. Find three points on this reflected graph.
▶ Show solution
(a) Reflection in x-axis: $y = -3^x$
Points: $(0,-1)$, $(1,-3)$, $(2,-9)$ — all y-values negated.
(b) Reflection in y-axis: $y = 3^{-x} = \left(\frac{1}{3}\right)^x$
Points: $(0,1)$, $(-1,3)$, $(-2,9)$ — all x-values negated.
Note: $(0,1)$ stays fixed in both reflections (it lies on both axes).
The graph of $y = f(x)$ has a maximum at $(2, 7)$ and crosses the x-axis at $x = -1$ and $x = 5$.
The graph of $y = f(x + 3) + 2$ is drawn. Find:
(a) The coordinates of the maximum on the new graph.
(b) The x-coordinates where the new graph crosses the x-axis.
▶ Show solution
The transformation is: left 3 (subtract 3 from x), up 2 (add 2 to y).
(a) Maximum: $(2, 7) \to (2-3,\ 7+2) = (-1,\ 9)$
(b) x-intercepts of $f(x)$ are at $x=-1$ and $x=5$, where $y=0$.
After transformation: x-values shift left by 3.
$x=-1 \to x=-4$; $x=5 \to x=2$.
But the y-values also shift up by 2, so $y=0$ on $f$ becomes $y=2$ on the new graph.
So the new graph does NOT cross the x-axis at these points — it crosses where $f(x+3) = -2$.
These cannot be found without more information about $f$. The new x-intercepts are not simply at $x=-4$ and $x=2$ — this is a common error.
$f(x) = x^2 - 4$.
(a) Sketch $y = f(x)$, labelling the roots and y-intercept.
(b) On the same axes, sketch $y = -f(x)$, labelling the turning point.
(c) On a new set of axes, sketch $y = f(x - 2)$, labelling the vertex and roots.
(d) Write down the equation of the graph produced when $y = f(x)$ is reflected in the y-axis. Is it different from $f(x)$?
▶ Show solution
(a) $y = x^2 - 4$: roots at $x = \pm 2$ (i.e. $(-2,0)$ and $(2,0)$); y-intercept at $(0,-4)$; vertex (minimum) at $(0,-4)$. U-shaped parabola.
(b) $y = -f(x) = -(x^2-4) = -x^2+4$: Same roots at $x = \pm 2$; turning point flips to maximum at $(0, 4)$. ∩-shaped parabola.
(c) $y = f(x-2) = (x-2)^2-4$: Shift right 2. Vertex moves from $(0,-4)$ to $(2,-4)$. New roots: $(x-2)^2 = 4 \Rightarrow x-2 = \pm 2 \Rightarrow x=0$ or $x=4$.
(d) $y = f(-x) = (-x)^2-4 = x^2-4$. This is identical to $f(x)$! Since $y=x^2-4$ is symmetric about the y-axis (it only contains even powers of $x$), reflecting it in the y-axis gives the same graph.