Horizontal Transformations

• Horizontal Shift: $f(x - h)$ shifts the graph right by $h$ units (left if $h < 0$)
• Horizontal Stretch/Compression: $f(bx)$
  • 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

1. Horizontal shifts and stretches (inside the function)
2. Vertical stretches and reflections (coefficient)
3. 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

• 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 $(-\infty, -3]$

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) ✓