Function Transformations

MATH161

Reference: Stewart 1.3 • Chapter: 1 • Section: 3

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Function Transformations

Why Learn to Transform Graphs?

Imagine you know what $y = x^2$ looks like. Now what if someone asks you to graph $y = 3(x-2)^2 + 5$? You could plot dozens of points... or you could recognize this as the same parabola, just moved and stretched. That's the power of transformations: graph complex functions by modifying simple ones you already know.

This skill is a visual shortcut. Once you master it, you can sketch graphs in seconds instead of minutes. More importantly, transformations reveal the structure hiding inside complicated-looking functions—structure that becomes essential when you study limits, derivatives, and differential equations.

Prerequisite Map

Quick Reference

Property	Value
Course	MATH 161
Chapter	1.3
Difficulty	Beginner
Time	~20 minutes

Key Concepts

The Transformation Toolkit

Starting from a parent function $y = f(x)$, here are all the ways to transform it:

Transformation	Formula	Effect
Vertical shift up	$y = f(x) + c$	Move graph up by $c$ units
Vertical shift down	$y = f(x) - c$	Move graph down by $c$ units
Horizontal shift right	$y = f(x - c)$	Move graph right by $c$ units
Horizontal shift left	$y = f(x + c)$	Move graph left by $c$ units
Vertical stretch	$y = cf(x)$ where $c > 1$	Stretch away from $x$-axis
Vertical shrink	$y = cf(x)$ where $0 < c < 1$	Compress toward $x$-axis
Horizontal shrink	$y = f(cx)$ where $c > 1$	Compress toward $y$-axis
Horizontal stretch	$y = f(cx)$ where $0 < c < 1$	Stretch away from $y$-axis
Reflect over $x$-axis	$y = -f(x)$	Flip upside down
Reflect over $y$-axis	$y = f(-x)$	Flip left-right
Absolute value	$y = \\|f(x)\\|$	Reflect negative parts upward

The "Opposite Sign" Rule for Horizontal Shifts

This is the most common source of errors. Horizontal transformations work "backwards" from what you might expect:

$$y = f(x - 3) \text{ shifts RIGHT by 3}$$ $$y = f(x + 3) \text{ shifts LEFT by 3}$$

Why? To get the same output, you need to input a value that's 3 units larger when you're subtracting 3 inside.

Visualizing Shifts

Vertical Shifts (add/subtract OUTSIDE):

    y = f(x) + 2    ↑ moves UP 2
         ●
        / \
    ───────────
        / \
         ●
    y = f(x) - 2    ↓ moves DOWN 2


Horizontal Shifts (add/subtract INSIDE):

y = f(x+2)  y = f(x)  y = f(x-2)
    ●           ●           ●
   / \         / \         / \
  ←             │           →
  LEFT 2        │         RIGHT 2

Order of Transformations

When multiple transformations are combined, apply them in this order:

Horizontal shifts (inside the function)
Stretches/shrinks and reflections
Vertical shifts (outside the function)

For $y = 2f(x - 3) + 1$:

First: shift right 3
Second: stretch vertically by factor 2
Third: shift up 1

Why This Works

Think of transformations as modifying the coordinate system rather than the function:

Vertical changes (outside $f$): directly change the output values
Horizontal changes (inside $f$): change what input is needed to get the same output

This is why horizontal transformations appear "reversed"—you're changing the input requirement.

Practice Problems

Level 1 Identifying a Single Transformation

Describe the transformation that takes $y = x^2$ to $y = x^2 + 4$.

Thought Process

Compare the two equations. The only difference is "$+ 4$" which is outside the function (added after squaring). Changes outside affect vertical position.

Show Answer

The graph of $y = x^2$ is shifted up by 4 units.

Adding a constant outside the function moves the graph vertically. Since we're adding a positive number, the graph moves up.

Level 2 Horizontal Shift Recognition

The graph of $y = \sqrt{x}$ is transformed to $y = \sqrt{x + 5}$. Describe the transformation and state the new domain.

Thought Process

The "$+5$" is inside the square root (affecting the input). Remember the opposite sign rule: adding inside means shifting left. For domain, set the expression inside the square root $\geq 0$.

Show Answer

Transformation: Shift left by 5 units.

New domain: We need $x + 5 \geq 0$, so $x \geq -5$.

Domain: $[-5, \infty)$

Note: The original domain of $\sqrt{x}$ was $[0, \infty)$. Shifting left by 5 moves this to $[-5, \infty)$.

Level 3 Multiple Transformations

Starting with $y = \vert x\vert $, describe the sequence of transformations needed to obtain $y = -2\vert x - 1\vert + 3$.

Thought Process

Break down the equation by identifying each modification to $\vert x\vert $:

Inside: $x - 1$ (horizontal shift)
The coefficient $-2$ (reflection and stretch)
The $+3$ (vertical shift)

Apply in the standard order: horizontal first, then stretches/reflections, then vertical.

Show Answer

Starting with $y = \vert x\vert $:

Shift right 1 unit: $y = \vert x - 1\vert $
Stretch vertically by factor 2: $y = 2\vert x - 1\vert $
Reflect over the $x$-axis: $y = -2\vert x - 1\vert $
Shift up 3 units: $y = -2\vert x - 1\vert + 3$

The final graph is an upside-down V-shape, twice as steep as $\vert x\vert $, with vertex at $(1, 3)$.

Level 4 Writing the Transformed Equation

The graph of $y = f(x)$ passes through the point $(2, 5)$. After a reflection over the $y$-axis followed by a horizontal stretch by a factor of 3, what point does the new graph pass through? Write the equation of the transformed function.

Thought Process

Work through transformations one at a time:

Reflection over $y$-axis: replace $x$ with $-x$, so $(2, 5) \to (-2, 5)$
Horizontal stretch by 3: replace $x$ with $x/3$, and the point's $x$-coordinate gets multiplied by 3

For the equation, apply these same replacements to the input of $f$.

Show Answer

Tracking the point:

Start: $(2, 5)$
Reflect over $y$-axis: $(-2, 5)$
Horizontal stretch by 3: $(-6, 5)$

The equation:

Reflect over $y$-axis: $y = f(-x)$
Horizontal stretch by 3: replace $x$ with $x/3$

$$y = f\left(-\frac{x}{3}\right)$$

Verification: When $x = -6$: $f(-(-6)/3) = f(2) = 5$ ✓

Level 5 Discovering Symmetry Through Transformations

A function $f$ is called even if $f(-x) = f(x)$ for all $x$ in its domain.

What does the transformation $y = f(-x)$ do geometrically?
Explain why even functions are symmetric about the $y$-axis.
If $g(x) = (x-2)^4 + (x-2)^2$, is $g$ symmetric about any vertical line? If so, which one?

Thought Process

Part (a): Recall what replacing $x$ with $-x$ does.

Part (b): If $f(-x) = f(x)$, then the original and reflected graphs are identical...

Part (c): Look for a substitution that reveals an even function structure. Let $u = x - 2$.

Show Answer

(a) The transformation $y = f(-x)$ reflects the graph of $f$ over the $y$-axis.

(b) If $f(-x) = f(x)$, then reflecting the graph over the $y$-axis produces the identical graph. The only way a shape can be unchanged by a reflection is if it's symmetric about the line of reflection. Therefore, even functions are symmetric about the $y$-axis.

(c) Let $u = x - 2$. Then: $$g(x) = u^4 + u^2$$

Define $h(u) = u^4 + u^2$. This is an even function because: $$h(-u) = (-u)^4 + (-u)^2 = u^4 + u^2 = h(u)$$

Since $h(u)$ is symmetric about $u = 0$, and $u = x - 2$, the function $g(x)$ is symmetric about the line $x = 2$.

Mastery Checklist

[ ] I can identify vertical and horizontal shifts from an equation
[ ] I apply the "opposite sign" rule correctly for horizontal shifts
[ ] I can distinguish between stretches and shrinks (vertical and horizontal)
[ ] I can identify reflections over the $x$-axis and $y$-axis
[ ] I can combine multiple transformations in the correct order
[ ] I can write the equation for a transformed function
[ ] I understand how transformations affect the domain

Mental Model

The Costume Change Analogy: Think of the parent function as an actor. Transformations are like costume changes:

Shifts move the actor to a new spot on stage
Stretches/shrinks change how tall or wide the costume is
Reflections flip the costume (mirror image)

The actor (the shape of the curve) is the same—only their position and appearance change. Once you recognize the actor, you can quickly describe any costume they're wearing.

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Last updated: 2026-01-22