\(\mathrm{h(x) = 2(x - 4)^2 - 32}\)The quadratic function h is defined as shown. In the xy-plane, the graph of...

1. TRANSLATE the problem information

• Given information:
  • \(\mathrm{h(x) = 2(x - 4)^2 - 32}\)
  • Graph intersects x-axis at \(\mathrm{(0, 0)}\) and \(\mathrm{(t, 0)}\)
  • Need to find the value of t

• What this tells us: X-intercepts occur when \(\mathrm{h(x) = 0}\)

2. INFER the solution approach

• To find x-intercepts, we need to solve \(\mathrm{h(x) = 0}\)
• Set up the equation: \(\mathrm{2(x - 4)^2 - 32 = 0}\)

3. SIMPLIFY through algebraic steps

• Start with: \(\mathrm{2(x - 4)^2 - 32 = 0}\)
• Add 32 to both sides: \(\mathrm{2(x - 4)^2 = 32}\)
• Divide both sides by 2: \(\mathrm{(x - 4)^2 = 16}\)
• Take square root of both sides: \(\mathrm{x - 4 = ±4}\)

4. CONSIDER ALL CASES from the square root

• Case 1: \(\mathrm{x - 4 = 4}\), so \(\mathrm{x = 8}\)
• Case 2: \(\mathrm{x - 4 = -4}\), so \(\mathrm{x = 0}\)

5. INFER the final answer

• The x-intercepts are at \(\mathrm{x = 0}\) and \(\mathrm{x = 8}\)
• Since the problem states intersections at \(\mathrm{(0, 0)}\) and \(\mathrm{(t, 0)}\), we have \(\mathrm{t = 8}\)

Answer: D. 8

Why Students Usually Falter on This Problem

Most Common Error Path:

Poor TRANSLATE skill: Students might not recognize that "intersects the x-axis" means \(\mathrm{h(x) = 0}\). Instead, they might try to substitute \(\mathrm{x = 0}\) and \(\mathrm{x = t}\) directly into the function without setting it equal to zero. This leads to confusion about what equation to solve and often results in random guessing.

Second Most Common Error:

Weak CONSIDER ALL CASES reasoning: After getting \(\mathrm{(x - 4)^2 = 16}\) and taking the square root, students might only consider the positive case (\(\mathrm{x - 4 = 4}\), giving \(\mathrm{x = 8}\)) and miss the negative case (\(\mathrm{x - 4 = -4}\), giving \(\mathrm{x = 0}\)). This could lead them to think there's an error since they can't verify both given intercept points, causing them to select Choice C (4) by incorrectly thinking the vertex x-coordinate is the answer.

The Bottom Line:

This problem tests whether students can properly translate the geometric concept of x-intercepts into the algebraic condition of setting the function equal to zero, then systematically solve the resulting equation while considering all possible solutions.