Function Transformations and Geometric Solutions

Understanding Function Notation

To master transformations, we must first be clear on what the notation $y = f(x)$ represents. This notation indicates that for any given value on the horizontal axis (the $x$ value), the corresponding vertical value (the $y$ value) on the curve is calculated using the rule defined by $f(x)$. For example, if we have the function $y = x^2 + 3$, the $y$ value located above $x = 2$ is $2^2 + 3 = 7$, and the value above $x = 4$ is $4^2 + 3 = 19$. We use this understanding of inputs and outputs to deduce how modifications to the algebra affect the geometry of the graph.

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

Consider the specific case where we compare $y = f(x) = x^3$ with $y = 4f(x) = 4x^3$. In this transformation, every $y$ value in the new function is exactly four times as large as the original $y$ value for the same $x$ coordinate. Geometrically, this is a vertical stretch of the graph by a factor of 4. The stretch occurs away from the $x$ axis: points move upwards if $y$ is positive and downwards if $y$ is negative.

If the constant $a$ is between 0 and 1, such as $a = 0.5$, the graph becomes half its original height. If $a$ is negative, the graph is stretched by the magnitude of $a$ and then reflected in the $x$ axis due to the change in sign.

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

When we compare $y = f(x) = x^2$ with $y = f(x) + 3 = x^2 + 3$, we find that every $y$ value increases by 3 units. This translates the entire graph upwards parallel to the $y$ axis. Formally, we describe this as a translation by the vector $\binom{0}{3}$. In general, the transformation $y = f(x) + a$ results in a translation by $\binom{0}{a}$.

In trigonometry, notation requires care. Writing $\cos x + a$ can be ambiguous, as it might be confused with $\cos(x + a)$. Mathematicians typically write $a + \cos x$ to clarify that the constant is added to the result of the cosine function, not to the angle itself. Similarly, $\cos^{-1} x$ refers to the inverse function (arccos) and not $1/\cos x$, which is written as $\sec x$.

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

This transformation is often counter-intuitive. Adding a positive constant $a$ to the input $x$ actually shifts the graph to the left, in the negative $x$ direction. To understand this, let $f(x) = 2^x$ and compare it with $f(x + 3)$. On the original graph, the $y$ value at $x = 5$ is $2^5 = 32$. On the new graph $y = f(x + 3)$, the $y$ value above $x = 2$ is $f(2 + 3) = f(5) = 32$. Thus, the value that used to be at $x = 5$ has moved to $x = 2$. This requires shifting the graph 3 units to the left.

In general, $y = f(x + a)$ is a translation of $y = f(x)$ by the vector $\binom{-a}{0}$. If $a$ is negative, such as $y = f(x - 4)$, the graph translates 4 units to the right.

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

1. If $f(x) = x^2 + 2x - 5$, then $f(x + 3) = (x + 3)^2 + 2(x + 3) - 5$.
2. If $f(x) = \cos(2x)$, then $f(x - \pi/2) = \cos(2(x - \pi/2)) = \cos(2x - \pi)$. Note that the coefficient 2 must multiply the entire bracketed replacement.

Horizontal Squash: $y = f(ax)$

Multiplying the input $x$ by a factor $a$ squashes the graph towards the $y$ axis by a factor of $a$. Consider $f(x) = x^3 - 3x^2 + 2$ and $f(2x)$. At $x = 1$, the function $f(2x)$ takes the value of $f(2)$, which is $-2$. This value originally occurred at $x = 2$, but now occurs at $x = 1$, effectively compressing the graph horizontally.

If $0 < a < 1$, the graph is stretched horizontally away from the $y$ axis. If $a$ is negative, the graph is reflected in the $y$ axis in addition to the horizontal scaling.

Summary of Transformations

Notation	Transformation Type
$af(x)$	Vertical stretch factor $a$ away from $x$ axis. Reflection in $x$ axis if $a < 0$.
$f(x) + a$	Translation by vector $\binom{0}{a}$.
$f(x + a)$	Translation by vector $\binom{-a}{0}$.
$f(ax)$	Horizontal squash factor $a$ towards $y$ axis. Reflection in $y$ axis if $a < 0$.

Combining Transformations

The order of operations is critical when multiple transformations are applied. Consider $y = \cos(2x + \pi/6)$. We could reach this via two different sequences:

1. Translate by $-\pi/6$ first: $\cos x \rightarrow \cos(x + \pi/6)$, then squash by factor 2: $\cos(2x + \pi/6)$. This is correct.
2. Squash by factor 2 first: $\cos x \rightarrow \cos 2x$, then translate by $-\pi/6$: $\cos 2(x + \pi/6) = \cos(2x + \pi/3)$. This is incorrect. Alternatively, if you squash by 2 first, you would need to translate by only $-\pi/12$ to reach $\cos 2(x + \pi/12) = \cos(2x + \pi/6)$.

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

The notation $f(g(x))$ means that the function $g(x)$ is the input for function $f$. If $g(x) = 2x$ and $f(x) = x^2 + 3x - 2$, we find $f(g(x))$ by replacing every $x$ in $f$ with $2x$: $f(g(x)) = (2x)^2 + 3(2x) - 2 = 4x^2 + 6x - 2$. Note that $f(g(x))$ is generally not equal to $g(f(x))$.

Linear and Quadratic Graphs as Transformations

Any linear graph $y = mx + c$ can be viewed as a series of transformations of $y = x$. One sequence is a vertical stretch of $y = x$ by factor $m$, followed by a translation of $c/m$ to the left: $y = x \rightarrow y = mx \rightarrow y = m(x + c/m) = mx + c$.

Similarly, a quadratic in the form $y = a(x + b)^2 + c$ is a transformation of $y = x^2$:

1. Vertical stretch factor $a$: $y = ax^2$.
2. Translation by $-b$ horizontally: $y = a(x + b)^2$.
3. Translation by $c$ vertically: $y = a(x + b)^2 + c$.

Geometric Interpretation of Equations

The solutions to the equation $f(x) = g(x)$ correspond to the $x$ coordinates of the points where the graphs of $y = f(x)$ and $y = g(x)$ intersect. If the equations are solved simultaneously, the resulting $(x, y)$ pairs represent the geometric locations where the two curves meet. If the graphs do not intersect, there are no real solutions to the simultaneous equations.