The function p is defined by \(\mathrm{p(x) = k(x+2)^2 - 7}\), where k is a constant. In the xy-plane, the...

1. TRANSLATE the transformation information

• Given information:
  • Original function: \(\mathrm{p(x) = k(x+2)^2 - 7}\)
  • Translation: 5 units left and 6 units up
  • Need to find: equation for \(\mathrm{q(x)}\)

2. INFER the transformation approach

• For any function \(\mathrm{f(x)}\):
  • To move h units left: replace x with \(\mathrm{(x+h)}\)
  • To move k units up: add k to the entire function
• Strategy: Apply both transformations to get \(\mathrm{q(x) = p(x+5) + 6}\)

3. SIMPLIFY by finding p(x+5)

• Start with \(\mathrm{p(x) = k(x+2)^2 - 7}\)
• Replace x with \(\mathrm{(x+5)}\): \(\mathrm{p(x+5) = k((x+5)+2)^2 - 7}\)
• Combine terms inside parentheses: \(\mathrm{p(x+5) = k(x+7)^2 - 7}\)

4. SIMPLIFY by adding the vertical translation

• \(\mathrm{q(x) = p(x+5) + 6}\)
• \(\mathrm{q(x) = k(x+7)^2 - 7 + 6}\)
• \(\mathrm{q(x) = k(x+7)^2 - 1}\)

Answer: C) \(\mathrm{q(x) = k(x+7)^2 - 1}\)

Why Students Usually Falter on This Problem

Most Common Error Path:

Weak TRANSLATE skill: Students confuse the direction of horizontal translations, thinking "5 units left" means replace x with \(\mathrm{(x-5)}\) instead of \(\mathrm{(x+5)}\).

This incorrect reasoning leads to \(\mathrm{q(x) = p(x-5) + 6 = k((x-5)+2)^2 - 1 = k(x-3)^2 - 1}\), causing them to select Choice B (\(\mathrm{k(x-3)^2 - 1}\)).

Second Most Common Error:

Poor SIMPLIFY execution: Students correctly identify the transformation but make algebraic errors when combining the vertical shifts, calculating \(\mathrm{-7 + 6 = -13}\) instead of \(\mathrm{-1}\).

This leads them to think \(\mathrm{q(x) = k(x+7)^2 - 13}\), causing them to select Choice D (\(\mathrm{k(x+7)^2 - 13}\)).

The Bottom Line:

Function transformations require careful attention to direction (especially for horizontal shifts, which work opposite to intuition) and precise algebraic manipulation when combining multiple shifts.