A projectile is launched upward, and its height \(\mathrm{h(t)}\), in meters, is modeled by a quadratic function of time t,...

1. TRANSLATE the problem information

• Given information:
  • Vertex is at (2, 10)
  • One zero at \(\mathrm{t = 1/4}\)
  • Need to find the other zero

2. INFER the key relationship

• Since we have a quadratic function (parabola), the zeros must be symmetric about the axis of symmetry
• The axis of symmetry passes through the vertex, so it's the vertical line \(\mathrm{t = 2}\)
• This symmetry property is the key to solving efficiently

3. SIMPLIFY to find the distance

• Distance from known zero to axis of symmetry:
  \(\mathrm{|2 - 1/4|}\)
  \(\mathrm{= |8/4 - 1/4|}\)
  \(\mathrm{= 7/4}\)

4. APPLY CONSTRAINTS using symmetry

• The other zero must be the same distance (7/4) on the opposite side of \(\mathrm{t = 2}\)
• Other zero:
  \(\mathrm{t = 2 + 7/4}\)
  \(\mathrm{= 8/4 + 7/4}\)
  \(\mathrm{= 15/4}\)

Answer: B. 15/4

Why Students Usually Falter on This Problem

Most Common Error Path:

Weak INFER skill: Students don't recognize the symmetry property of parabolic roots and instead try to set up the full quadratic equation using vertex form.

They might write \(\mathrm{h(t) = a(t - 2)^2 + 10}\), substitute the known zero to solve for 'a', then solve the resulting equation. While this approach works, it's much more complex and creates opportunities for algebraic errors. This lengthy approach may cause them to run out of time or make calculation mistakes, leading to confusion and guessing.

Second Most Common Error:

Poor TRANSLATE reasoning: Students misinterpret what "vertex at (2, 10)" means for the axis of symmetry.

They might think the axis of symmetry is at \(\mathrm{t = 10}\) instead of \(\mathrm{t = 2}\), or confuse which coordinate represents the axis. This leads to incorrect distance calculations and wrong final answers. This may lead them to select Choice A (7/4) if they use the y-coordinate as the axis.

The Bottom Line:

This problem tests whether students can recognize and apply the fundamental symmetry property of parabolas rather than getting bogged down in complex algebraic manipulations. The key insight is that quadratic functions have symmetric roots - once you see this, the solution becomes straightforward arithmetic.