Function Transformations

Understanding how to shift, stretch, compress, and reflect functions

Function Transformations

Types of Transformations

Given a parent function $f(x)$ , we can transform it in several ways:

Vertical Transformations

Vertical Shift: $f(x) + k$ shifts the graph up by $k$ units (down if $k < 0$ )
Vertical Stretch/Compression: $a \cdot f(x)$ $a \cdot f (x)$
- If $|a| > 1$ : vertical stretch
- If $0 < |a| < 1$ : vertical compression
- If $a < 0$ : reflection across x-axis

Horizontal Transformations

Horizontal Shift: $f(x - h)$ shifts the graph right by $h$ units (left if $h < 0$ )
Horizontal Stretch/Compression: $f(bx)$ $f (b x)$
- If $|b| > 1$ : horizontal compression (faster)
- If $0 < |b| < 1$ : horizontal stretch (slower)
- If $b < 0$ : reflection across y-axis

General Form

$g(x) = a \cdot f(b(x - h)) + k$

Where:

$a$ : vertical stretch/compression and reflection
$b$ : horizontal stretch/compression and reflection
$h$ : horizontal shift
$k$ : vertical shift

Order of Transformations

Horizontal shifts and stretches (inside the function)
Vertical stretches and reflections (coefficient)
Vertical shifts (outside)

Key Points to Remember

Inside changes ( $x$ transformations) work opposite to intuition
$f(x - 3)$ shifts RIGHT, not left
$f(2x)$ compresses horizontally, not stretches

📚 Practice Problems

1Problem 1easy

❓ Question:

The graph of $f(x) = x^2$ is transformed to $g(x) = 2(x-3)^2 + 1$ . Describe all transformations applied.

💡 Show Solution

Identify each transformation:

Starting with $f(x) = x^2$ , we have $g(x) = 2(x-3)^2 + 1$

Compare to $g(x) = a \cdot f(b(x - h)) + k$ :

$a = 2$ : vertical stretch by factor of 2
$b = 1$ : no horizontal stretch
$h = 3$ : horizontal shift right 3 units
$k = 1$ : vertical shift up 1 unit

Transformations in order:

Shift right 3 units
Stretch vertically by factor of 2
Shift up 1 unit

Vertex: The vertex of $f(x) = x^2$ is at $(0, 0)$ . After transformations, the new vertex is at $(3, 1)$ .

2Problem 2medium

❓ Question:

Given $f(x) = \sqrt{x}$ , write the equation for the function that results from: reflecting across the x-axis, shifting left 2 units, and shifting down 3 units.

💡 Show Solution

Apply transformations step by step:

Starting with $f(x) = \sqrt{x}$

Step 1: Reflect across x-axis

Multiply by -1: $-f(x) = -\sqrt{x}$

Step 2: Shift left 2 units

Replace $x$ with $x + 2$ : $-f(x+2) = -\sqrt{x+2}$

Step 3: Shift down 3 units

Subtract 3: $-f(x+2) - 3 = -\sqrt{x+2} - 3$

Final answer: $g(x) = -\sqrt{x+2} - 3$

Domain: Since we need $x + 2 \geq 0$ , the domain is $x \geq -2$ or $[-2, \infty)$

Range: Since $\sqrt{x+2} \geq 0$ , we have $-\sqrt{x+2} \leq 0$ , so $-\sqrt{x+2} - 3 \leq -3$ . Range is $(-\infty, -3]$

3Problem 3hard

❓ Question:

The point $(4, -2)$ lies on the graph of $y = f(x)$ . What point must lie on the graph of $y = -3f(\frac{1}{2}x + 1) + 5$ ?

💡 Show Solution

Work backwards from the transformed point:

We know $(4, -2)$ is on $y = f(x)$ , so $f(4) = -2$ .

For $g(x) = -3f(\frac{1}{2}x + 1) + 5$ , we need to find which $x$ makes the inside equal to 4.

Find the x-coordinate: Set $\frac{1}{2}x + 1 = 4$

Solve for $x$ : $\frac{1}{2}x = 3$ $x = 6$

Find the y-coordinate: When $x = 6$ : $g(6) = -3f(\frac{1}{2}(6) + 1) + 5$ $= -3f(4) + 5$ $= -3(-2) + 5$ $= 6 + 5 = 11$

Answer: The point $(6, 11)$ must lie on $y = -3f(\frac{1}{2}x + 1) + 5$

Verification of transformations:

Original point: $(4, -2)$
Horizontal: $\frac{1}{2}x + 1 = 4$ means shift left 2 then stretch by 2: $4 \to 2 \to 6$
Vertical: $-2 \to 6$ (multiply by -3) $\to 11$ (add 5) ✓

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