The function h is defined by \(\mathrm{h(x) = a^x + b}\), where a and b are positive constants. The graph...

1. TRANSLATE the given information into mathematical equations

• Given information:
  • Function: \(\mathrm{h(x) = a^x + b}\) (a, b positive constants)
  • Point (0, 10): When \(\mathrm{x = 0}\), \(\mathrm{h(x) = 10}\)
  • Point (-2, 325/36): When \(\mathrm{x = -2}\), \(\mathrm{h(x) = \frac{325}{36}}\)
• What this tells us: We can substitute these points into our function to create two equations with unknowns a and b.

2. INFER the solution strategy

• Since we have two unknowns (a and b) and two points, we can create a system of equations
• Strategy: Substitute each point into \(\mathrm{h(x) = a^x + b}\) and solve for the constants

3. SIMPLIFY using the first point (0, 10)

• Substitute \(\mathrm{x = 0}\), \(\mathrm{h(x) = 10}\) into \(\mathrm{h(x) = a^x + b}\):
  • \(\mathrm{h(0) = a^0 + b = 10}\)
  • Since \(\mathrm{a^0 = 1}\): \(\mathrm{1 + b = 10}\)
  • Therefore: \(\mathrm{b = 9}\)

4. SIMPLIFY using the second point (-2, 325/36)

• Substitute \(\mathrm{x = -2}\), \(\mathrm{h(x) = \frac{325}{36}}\), and \(\mathrm{b = 9}\):
  • \(\mathrm{h(-2) = a^{-2} + 9 = \frac{325}{36}}\)
  • Since \(\mathrm{a^{-2} = \frac{1}{a^2}}\): \(\mathrm{\frac{1}{a^2} + 9 = \frac{325}{36}}\)
  • Subtract 9 from both sides: \(\mathrm{\frac{1}{a^2} = \frac{325}{36} - 9}\)
  • Convert 9 to same denominator: \(\mathrm{\frac{1}{a^2} = \frac{325}{36} - \frac{324}{36} = \frac{1}{36}}\)
  • Therefore: \(\mathrm{a^2 = 36}\)

5. APPLY CONSTRAINTS to find the final values

• Taking the square root: \(\mathrm{a = \pm 6}\)
• Since the problem states a is positive: \(\mathrm{a = 6}\)
• Calculate ab: \(\mathrm{ab = 6 \times 9 = 54}\)

Answer: C. 54

Why Students Usually Falter on This Problem

Most Common Error Path:

Weak TRANSLATE skill: Students struggle to correctly substitute negative exponents or forget the zero exponent rule.

Many students incorrectly evaluate \(\mathrm{a^0}\) as 0 instead of 1, or they get confused by the negative exponent in \(\mathrm{a^{-2}}\). This leads to completely wrong equations from the start, making it impossible to solve correctly. This leads to confusion and guessing among the answer choices.

Second Most Common Error:

Poor SIMPLIFY execution: Students make arithmetic errors when working with fractions, particularly when subtracting 9 from 325/36.

The fraction arithmetic step (\(\mathrm{\frac{325}{36} - 9 = \frac{325}{36} - \frac{324}{36}}\)) requires careful attention to common denominators. Students who rush this step or make computational errors will get incorrect values for a, leading them to select Choice A (1/4) or Choice B (1/2).

The Bottom Line:

This problem tests whether students can systematically use point conditions with exponential functions while maintaining accuracy through multiple algebraic steps. The combination of exponent rules and fraction arithmetic creates multiple opportunities for error, making careful execution essential.