Transformations and Compositions of Functions

Introduction to Transformations

Graph transformation is a topic that is often poorly understood because students frequently learn rules without grasping the underlying mechanics. It is particularly tricky because horizontal shifts like $y = f(x + a)$ appear as though they should shift a graph to the right, yet they actually move it to the left when $a$ is positive. To truly master this topic for the TMUA, you must understand how the notation $y = f(x)$ identifies the $y$ value above a given $x$ coordinate as the result of a calculation, and how modifications to that calculation affect the resulting curve.

Vertical Stretch: $y = af(x)$

In the transformation $y = af(x)$, every original $y$ value is multiplied by the factor $a$. For example, if we compare $y = f(x) = x^3$ with $y = 4f(x) = 4x^3$, each $y$ value is four times as large. This corresponds to a vertical stretch away from the $x$ axis by a factor of $a$.

If $0 < a < 1$, the graph becomes less tall (a compression towards the $x$ axis). If $a$ is negative, the graph is reflected in the $x$ axis in addition to the vertical stretch.

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

Adding a constant $a$ to the function result, as in $y = f(x) + a$, shifts every point on the graph up by $a$ units. This is a translation by the vector $\binom{0}{a}$. For instance, moving from $y = x^2$ to $y = x^2 + 3$ shifts the entire parabola up by 3 units parallel to the $y$ axis.

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

This transformation shifts the graph parallel to the $x$ axis. To find the expression for $f(x+a)$ given $f(x)$, replace every $x$ in the original function with $(x + a)$.

Worked Example 1 Given $f(x) = x^2 + 2x - 5$, find an expression for $f(x + 3)$. Substituting $(x+3)$ for $x$: $f(x + 3) = (x + 3)^2 + 2(x + 3) - 5$.

Worked Example 2 Given $f(x) = \cos(2x)$, find an expression for $f(x - \frac{\pi}{2})$. Replacing $x$ with $(x - \frac{\pi}{2})$: $f(x - \frac{\pi}{2}) = \cos 2(x - \frac{\pi}{2}) = \cos(2x - \pi)$.

Note that the 2 multiplies the entire replacement: we must have $\cos 2(x - \frac{\pi}{2})$ and not $\cos(2x - \frac{\pi}{2})$.

Understanding the Shift Consider $f(x) = 2^x$. In $y = f(x)$, $x = 2$ gives $y = 4$. In $y = f(x + 3)$, $x = 2$ gives $y = f(2 + 3) = f(5) = 32$. The $y$ value found at $x = 5$ on the original graph now appears at $x = 2$ on the new graph. This means the graph of $y = f(x + a)$ is the graph of $y = f(x)$ translated by $\binom{-a}{0}$. If $a$ is positive, it moves left.

Horizontal Squash: $y = f(ax)$

In $y = f(ax)$, the $y$ value above a given $x$ is the value that would have occurred at $ax$ in the original function. If $a = 2$, the $y$ values occur at half the $x$ distance from the $y$ axis, effectively 'squashing' the graph towards the $y$ axis by a factor of $a$.

Worked Example 3 Given $f(x) = x^2$, find $f(3x)$. Replacing $x$ by $3x$: $f(3x) = (3x)^2 = 9x^2$.

Worked Example 4 Given $f(x) = \cos(2x + 30)$, find $f(4x)$. Replacing $x$ by $4x$: $f(4x) = \cos(2(4x) + 30) = \cos(8x + 30)$.

Summary of Transformations

Transformation	Effect
$af(x)$	Vertical stretch away from the $x$ axis by factor $a$. If $a < 0$, includes reflection in $x$ axis.
$f(x) + a$	Translation by $\binom{0}{a}$ (Vertical shift).
$f(x + a)$	Translation by $\binom{-a}{0}$ (Horizontal shift).
$f(ax)$	Horizontal squash towards the $y$ axis by factor $a$. If $a < 0$, includes reflection in $y$ axis.

Combining Transformations and Composite Functions

Order matters when combining transformations. Consider $y = \cos(2x + \frac{\pi}{6})$. If we translate by $-\frac{\pi}{6}$ then squash by factor 2, we get: $\cos(x) \rightarrow \cos(x + \frac{\pi}{6}) \rightarrow \cos(2x + \frac{\pi}{6})$. This is correct. However, squashing by 2 then translating by $-\frac{\pi}{6}$ gives: $\cos(x) \rightarrow \cos(2x) \rightarrow \cos(2(x + \frac{\pi}{6})) = \cos(2x + \frac{\pi}{3})$. This is incorrect.

Composite Function Notation $f(g(x))$ The notation $f(g(x))$ means taking the output of $g(x)$ and using it as the input for $f(x)$.

Worked Example 5 Let $g(x) = 2x$ and $f(x) = x^2 + 3x - 2$. To find $f(g(x))$, replace the input $x$ in $f(x)$ with $2x$: $f(g(x)) = (2x)^2 + 3(2x) - 2 = 4x^2 + 6x - 2$.