Function Transformations

6 min read

Reference: Stewart §1.3

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.

See It: Build the Transformation

The dashed curve is $y = 3(x-2)^2 + 5$. Start from the parent parabola and turn the knobs: stretch it, slide it sideways, slide it up. Line your solid curve up with the dashed target.

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.

Common misconception

$f(x+3)$ shifts the graph right because “$+3$ means moving right.”

This is the input-output-confusion error. The “$+3$” is inside the function, acting on the INPUT, not the output. The graph of $f(x+3)$ reaches the same height at $x$ as $f$ reaches at $x+3$. To find where the peak of $f(x+3)$ lands, ask: what $x$ makes the input equal what it was before? If the peak of $f$ was at $x_0$, then $f(x+3)$ peaks where $x+3 = x_0$, which means $x = x_0 - 3$ -- three units to the LEFT. A concrete check: $f(x) = x^2$ has vertex at $x=0$. Then $f(x+3) = (x+3)^2$ has vertex where $x+3=0$, i.e., $x=-3$. The vertex moved LEFT by 3, not right.

Common misconception

reading a graph’s shape to infer whether a transformation “looks right.”

This is the iconic-graph error applied to transformations. Students sometimes sketch a shifted parabola and judge the direction by how the sketch “feels” rather than by checking a specific point. The reliable check is to pick one known point on $f$ -- say the vertex -- compute where it lands after the transformation, and verify. A graph that “looks shifted right” to an untrained eye may actually shift left. Trust the algebra: find the new location of one anchor point, then draw.

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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Function Representations	Section 1.3	Function Arithmetic

Last updated: 2026-01-22