\(\mathrm{h(x) = \log_{2}(x - 3) + k}\), where k is a constant. In the given function h, the graph in...

1. TRANSLATE the transformation description

• Given information:
  • Original function: \(\mathrm{h(x) = log_2(x - 3) + k}\)
  • Transformed 2 units up and 5 units to the left
  • Need to find j(x)

2. INFER the transformation approach

• Horizontal transformations affect the input (x-value inside the function)
• Vertical transformations affect the output (add/subtract to entire function)
• Apply transformations step by step for clarity

3. Apply horizontal transformation first

• Moving 5 units to the left means replace x with (x + 5)
• \(\mathrm{h(x) = log_2(x - 3) + k}\) becomes:
• \(\mathrm{log_2((x + 5) - 3) + k = log_2(x + 2) + k}\)

4. Apply vertical transformation

• Moving 2 units up means add 2 to the entire function
• \(\mathrm{j(x) = log_2(x + 2) + k + 2}\)

Answer: B

Why Students Usually Falter on This Problem

Most Common Error Path:

Weak INFER skill: Confusing the direction of horizontal transformations

Students often think "5 units to the left" means subtract 5 from x, leading to:
\(\mathrm{log_2((x - 5) - 3) + k + 2 = log_2(x - 8) + k + 2}\)

This may lead them to select Choice D (\(\mathrm{j(x) = log_2(x - 8) + k + 2}\))

Second Most Common Error:

Poor TRANSLATE reasoning: Misunderstanding vertical transformation direction

Students might interpret "2 units up" as subtracting 2, thinking it moves the graph down from their perspective, leading to:
\(\mathrm{j(x) = log_2(x + 2) + k - 2}\)

This may lead them to select Choice A (\(\mathrm{j(x) = log_2(x + 2) + k - 2}\))

The Bottom Line:

The key challenge is remembering that horizontal transformations work opposite to intuition (left means add, right means subtract), while vertical transformations work as expected (up means add, down means subtract).