Transformations of Functions for the ESAT

Understanding the Notation $y = f(x)$

Before exploring transformations, it is vital to be clear on what $y = f(x)$ represents. This notation tells us that the $y$ value directly above any given $x$ value is calculated using the rule $f(x)$. For example, if we have $y = x^2 + 3$, then for $x = 2$, we calculate $y = 2^2 + 3 = 7$. Similarly, for $x = 4$, we find $y = 19$. We use this principle to deduce how transformed graphs relate to the original.

Vertical Stretches: $y = a f(x)$

Let us compare $y = f(x) = x^3$ with $y = 4f(x) = 4x^3$. In this case, every $y$ value for the second function is four times as large as the $y$ value for the first function at the same $x$ coordinate. Geometrically, this represents a vertical stretch by a factor of 4. The stretching occurs away from the $x$ axis: points with positive $y$ values move up, and points with negative $y$ values move down.

In general, $y = af(x)$ is a vertical stretch parallel to the $y$ axis by a scale factor of $a$. Here is the general effect:

If $0 < a < 1$, the graph becomes less tall (it is effectively compressed vertically). If $a$ is negative, the graph is stretched by a factor of $|a|$ and reflected in the $x$ axis.

Vertical Translations: $y = f(x) + a$

Consider $y = f(x) = x^2$ and $y = f(x) + 3 = x^2 + 3$. In this transformation, every $y$ value increases by 3 units. This shifts the entire graph upwards parallel to the $y$ axis. Formally, we say the graph of $y = f(x)$ is translated by the vector $\binom{0}{a}$.

Note that in trigonometry, we often write $a + \cos x$ instead of $\cos x + a$ to avoid ambiguity, as $\cos x + a$ might be confused with $\cos(x + a)$. It is also important to remember that $\cos^{-1} x$ refers to the inverse function (sometimes called $\arccos x$) and not $\frac{1}{\cos x}$, which is actually $\sec x$.

Horizontal Translations: $y = f(x + a)$

This transformation is frequently misunderstood. Intuition might suggest that adding $a$ to $x$ should shift the graph to the right, but the opposite is true: for positive $a$, the graph shifts to the left. To understand why, we must look at how we calculate the expression and how the $y$ values align.

Working out the expression

To find $f(x + a)$, replace every $x$ in the original function with $(x + a)$.

Example 1: Given $f(x) = x^2 + 2x - 5$, find $f(x + 3)$. We replace $x$ with $x + 3$: $f(x + 3) = (x + 3)^2 + 2(x + 3) - 5$

Example 2: Given $f(x) = \cos(2x)$, find $f(x - \frac{\pi}{2})$. We replace $x$ with $x - \frac{\pi}{2}$: $f(x - \frac{\pi}{2}) = \cos 2(x - \frac{\pi}{2}) = \cos(2x - \pi)$ Note that the factor of 2 multiplies the entire replacement for $x$.

Why the shift is reversed

Consider $f(x) = 2^x$. On the original graph, when $x = 5$, $y = 32$. On the graph of $y = f(x + 3)$, the $y$ value above $x = 2$ is calculated as $f(2 + 3) = f(5) = 32$. This means the $y$ value that used to be at $x = 5$ has moved to $x = 2$. Consequently, the graph must shift 3 units to the left.

In general, $y = f(x + a)$ is a translation by $\binom{-a}{0}$.

Horizontal Squashes: $y = f(ax)$

To find the expression for $f(ax)$, replace every $x$ in $f(x)$ with $(ax)$. For example, if $f(x) = x^2$, then $f(3x) = (3x)^2 = 9x^2$. Note that this is not $3x^2$.

Geometrically, $y = f(ax)$ squashes the graph towards the $y$ axis by a factor of $a$. If $f(x) = x^3 - 3x^2 + 2$, then the $y$ value found at $x = 2$ in the original function will now be found at $x = 1$ in $f(2x)$, because $f(2 \times 1) = f(2)$.

If $0 < a < 1$, the graph is stretched horizontally away from the $y$ axis. If $a < 0$, there is a reflection in the $y$ axis.

Summary Table of Transformations

Transformation	Geometric Effect
$af(x)$	Vertical stretch parallel to $y$ axis by factor $a$. Reflection in $x$ axis if $a < 0$.
$f(x) + a$	Translation by $\binom{0}{a}$.
$f(x + a)$	Translation by $\binom{-a}{0}$.
$f(ax)$	Horizontal squash parallel to $x$ axis by factor $a$. Reflection in $y$ axis if $a < 0$.

Combining Transformations and Order

When combining transformations, the order in which you apply them matters significantly. Consider transforming $y = \cos x$ into $y = \cos(2x + \frac{\pi}{6})$. There are two ways to achieve this correctly:

1. Translate then Squash: Translate by $\binom{-\pi/6}{0}$ to get $\cos(x + \frac{\pi}{6})$, then squash horizontally by factor 2 to get $\cos(2x + \frac{\pi}{6})$.
2. Squash then Translate: Squash horizontally by factor 2 to get $\cos 2x$, then translate by $\binom{-\pi/12}{0}$. This works because replacing $x$ with $(x + \frac{\pi}{12})$ in $\cos 2x$ gives $\cos 2(x + \frac{\pi}{12}) = \cos(2x + \frac{\pi}{6})$.

Applying a squash by 2 followed by a translation of $\frac{\pi}{6}$ would result in $\cos 2(x + \frac{\pi}{6}) = \cos(2x + \frac{\pi}{3})$, which is incorrect.

Composite Functions: $f(g(x))$

The notation $f(g(x))$ means that $g(x)$ is used as the input for function $f$. If $g(x) = 2x$ and $f(x) = x^2 + 3x - 2$, then $f(g(x)) = (2x)^2 + 3(2x) - 2 = 4x^2 + 6x - 2$. In general, $f(g(x))$ is not equal to $g(f(x))$.

Applications to Lines and Quadratics

We can interpret $y = mx + c$ as a sequence of transformations of $y = x$. One way is to stretch by $m$ (to get $y = mx$) and then translate by $c$ (to get $y = mx + c$).

Similarly, $y = a(x + b)^2 + c$ is a transformation of $y = x^2$. We can apply a vertical stretch by $a$, then a horizontal translation by $-b$, and finally a vertical translation by $c$.