The function h is defined for real numbers x and takes positive values. Selected values of h are shown in...

1. TRANSLATE the problem information

• Given information:
  • Table showing x and h(x) values: \((-2,18)\), \((0,6)\), \((2,2)\), \((4,2/3)\), \((6,2/9)\)
  • Five exponential equation options to test
• We need to find which equation produces all these coordinate pairs

2. INFER the testing strategy

• Since we have specific points and multiple choice options, the most efficient approach is to test each equation systematically
• Start with a key point like \(\mathrm{x = 0}\) since any base to the power 0 equals 1, making calculations simpler
• If an equation fails any point, eliminate it immediately

3. SIMPLIFY by testing each option

Testing Option (A): \(\mathrm{h(x) = 6 \cdot 3^{x/2}}\)

\(\mathrm{h(0) = 6 \cdot 3^{0/2}}\)
\(\mathrm{= 6 \cdot 3^0}\)
\(\mathrm{= 6 \cdot 1}\)
\(\mathrm{= 6}\) ✓

\(\mathrm{h(2) = 6 \cdot 3^{2/2}}\)
\(\mathrm{= 6 \cdot 3^1}\)
\(\mathrm{= 6 \cdot 3}\)
\(\mathrm{= 18}\) ✗

Since \(\mathrm{h(2)}\) should be 2, not 18, Option A is wrong.

Testing Option (B): \(\mathrm{h(x) = 6 \cdot 3^{-x/2}}\)

\(\mathrm{h(0) = 6 \cdot 3^{0/2}}\)
\(\mathrm{= 6 \cdot 3^0}\)
\(\mathrm{= 6 \cdot 1}\)
\(\mathrm{= 6}\) ✓

\(\mathrm{h(2) = 6 \cdot 3^{-2/2}}\)
\(\mathrm{= 6 \cdot 3^{-1}}\)
\(\mathrm{= 6 \cdot (1/3)}\)
\(\mathrm{= 2}\) ✓

\(\mathrm{h(-2) = 6 \cdot 3^{-(-2)/2}}\)
\(\mathrm{= 6 \cdot 3^1}\)
\(\mathrm{= 6 \cdot 3}\)
\(\mathrm{= 18}\) ✓

\(\mathrm{h(4) = 6 \cdot 3^{-4/2}}\)
\(\mathrm{= 6 \cdot 3^{-2}}\)
\(\mathrm{= 6 \cdot (1/9)}\)
\(\mathrm{= 6/9}\)
\(\mathrm{= 2/3}\) ✓

\(\mathrm{h(6) = 6 \cdot 3^{-6/2}}\)
\(\mathrm{= 6 \cdot 3^{-3}}\)
\(\mathrm{= 6 \cdot (1/27)}\)
\(\mathrm{= 6/27}\)
\(\mathrm{= 2/9}\) ✓

All points match! But let's verify the remaining options aren't also correct.

Testing Option (C): \(\mathrm{h(x) = 6 \cdot (1/3)^x}\)

\(\mathrm{h(0) = 6 \cdot (1/3)^0}\)
\(\mathrm{= 6 \cdot 1}\)
\(\mathrm{= 6}\) ✓

\(\mathrm{h(2) = 6 \cdot (1/3)^2}\)
\(\mathrm{= 6 \cdot (1/9)}\)
\(\mathrm{= 6/9}\)
\(\mathrm{= 2/3}\) ✗

Since \(\mathrm{h(2)}\) should be 2, not \(\mathrm{2/3}\), Option C is wrong.

4. APPLY CONSTRAINTS to confirm our answer

• Only Option B satisfies all the given points
• The other options can be eliminated after failing just one or two test points

Answer: B

Why Students Usually Falter on This Problem

Most Common Error Path:

Weak SIMPLIFY execution: Students make calculation errors when working with negative exponents or fractional exponents, especially with expressions like \(\mathrm{3^{-x/2}}\).

For example, they might incorrectly calculate \(\mathrm{h(2) = 6 \cdot 3^{-2/2}}\) as \(\mathrm{6 \cdot 3^{-1} = 6 \cdot (-1/3) = -2}\) instead of \(\mathrm{6 \cdot (1/3) = 2}\). These sign errors or misunderstanding of negative exponents lead to eliminating the correct answer and potentially selecting Choice A (\(\mathrm{6 \cdot 3^{x/2}}\)) or getting confused and guessing.

Second Most Common Error:

Poor TRANSLATE reasoning: Students might not systematically test each option or might stop after finding that one option works for just the first point they check.

They might see that \(\mathrm{h(0) = 6}\) works for multiple options and not continue testing other points. This incomplete verification approach can lead them to select Choice A (\(\mathrm{6 \cdot 3^{x/2}}\)) if they only check \(\mathrm{x = 0}\), or causes them to get stuck and randomly select an answer.

The Bottom Line:

Success requires both systematic testing of all options AND accurate computation with exponential expressions. This problem rewards methodical verification over shortcuts.