Graph Transformations and Composite Functions for the ESAT

Understanding the notation $y = f(x)$ is the foundation for mastering graph transformations. This notation states that the $y$ value above any given $x$ value is calculated using the rule $f(x)$. For example, if $y = x^2 + 3$, the $y$ value for $x = 2$ is $2^2 + 3 = 7$. By applying modifications to this rule, we can deduce how the resulting graph relates to the original function.

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

Consider the transformation $y = a f(x)$. If we take $f(x) = x^3$ and $a = 4$, we compare $y = x^3$ with $y = 4x^3$. Every $y$ value in the new function is exactly four times as large as the corresponding $y$ value in the original. Geometrically, this is a vertical stretch away from the $x$ axis by a factor of 4. If $y$ is positive, the point moves further up: if $y$ is negative, the point moves further down.

In general, the graph of $y = af(x)$ is a vertical stretch of $y = f(x)$ parallel to the $y$ axis by scale factor $a$. If $0 < a < 1$, the graph becomes less tall (it squashes towards the $x$ axis). If $a$ is negative, the graph is reflected in the $x$ axis and stretched by a factor of $|a|$.

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

If we take $f(x) = x^2$ and $a = 3$, we compare $y = x^2$ with $y = x^2 + 3$. In this case, every $y$ value increases by 3 units. The entire graph shifts upwards parallel to the $y$ axis. Formally, we describe this as a translation by the vector $\binom{0}{3}$.

In general, $y = f(x) + a$ represents a translation of $y = f(x)$ by the vector $\binom{0}{a}$. If $a$ is negative, the graph shifts downwards. In trigonometry, note that $a + \cos x$ is often written instead of $\cos x + a$ to avoid ambiguity with $\cos(x + a)$.

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

This transformation is often misunderstood. Students frequently assume that adding $a$ to $x$ shifts the graph to the right, but the opposite is true: $y = f(x + a)$ shifts the graph to the left when $a$ is positive. To find the expression for $f(x+a)$, every $x$ in the original expression must be replaced by $(x+a)$.

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

To understand the shift, consider $f(x) = 2^x$. At $x=5$ on $y=f(x)$, the $y$ value is $2^5 = 32$. On the graph $y = f(x+3)$, we get this same $y$ value of 32 when $x=2$ because $f(2+3) = f(5)$. Thus, the point that was at $x=5$ has moved to $x=2$, which is 3 units to the left.

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

Horizontal Stretches: $y = f(ax)$

To find the expression for $f(ax)$, we replace every $x$ with $ax$. For example, if $f(x) = x^2$, then $f(3x) = (3x)^2 = 9x^2$. If $f(x) = \cos(2x + 30)$, then $f(4x) = \cos(2(4x) + 30) = \cos(8x + 30)$.

Geometrically, $y = f(ax)$ squashes the graph towards the $y$ axis by a factor of $a$. Consider $f(x) = x^3 - 3x^2 + 2$. On $y=f(x)$, the value $y=2$ occurs at $x=0$. On $y=f(2x)$, the $y$ value at $x=1$ is $f(2 \times 1) = f(2) = -2$, which is the $y$ value that originally occurred at $x=2$. The graph has been squashed horizontally by a factor of 2.

In general, $y = f(ax)$ is a horizontal stretch parallel to the $x$ axis by a scale factor of $\frac{1}{a}$. If $a < 0$, the graph is also reflected in the $y$ axis.

Composing Transformations

When multiple transformations are applied, the order of operations is critical. Consider transforming $y = \cos x$ into $y = \cos(2x + \frac{\pi}{6})$. There are two ways to achieve this correctly:

1. Translate by $\binom{-\frac{\pi}{6}}{0}$ first, then squash horizontally by factor 2: $\cos x \rightarrow \cos(x + \frac{\pi}{6}) \rightarrow \cos(2x + \frac{\pi}{6})$.

2. Squash horizontally by factor 2 first, then translate by $\binom{-\frac{\pi}{12}}{0}$: $\cos x \rightarrow \cos 2x \rightarrow \cos 2(x + \frac{\pi}{12}) = \cos(2x + \frac{\pi}{6})$.

Note that squashing by factor 2 first and then translating by $\frac{\pi}{6}$ would result in $\cos 2(x + \frac{\pi}{6}) = \cos(2x + \frac{\pi}{3})$, which is incorrect.

Composite Function Notation: $f(g(x))$

The notation $f(g(x))$ means that the output of $g(x)$ becomes the input for $f(x)$.

Example: If $g(x) = 2x$ and $f(x) = x^2 + 3x - 2$, then to find $f(g(x))$, we replace every $x$ in $f$ with $g(x)$: $f(g(x)) = (2x)^2 + 3(2x) - 2 = 4x^2 + 6x - 2$.

It is generally not true that $f(g(x)) = g(f(x))$. For instance, if $f(x) = x^2$ and $g(x) = x - 3$, then $f(g(x)) = (x-3)^2$ while $g(f(x)) = x^2 - 3$.