Question:A designer is creating a parabolic arch for a building entrance. The height function is modeled by \(\mathrm{f(x) = -8x^2...
GMAT Advanced Math : (Adv_Math) Questions
A designer is creating a parabolic arch for a building entrance. The height function is modeled by \(\mathrm{f(x) = -8x^2 + 16x + k}\), where \(\mathrm{x}\) represents the horizontal distance in meters and \(\mathrm{k}\) is a constant. For the arch to have the desired aesthetic, its graph must touch the ground (\(\mathrm{x}\)-axis) at exactly one point. What is the value of \(\mathrm{k}\)?
1. TRANSLATE the problem requirements
- Given information:
- Height function: \(\mathrm{f(x) = -8x^2 + 16x + k}\)
- The arch must "touch the ground at exactly one point"
- What this means mathematically: The parabola intersects the x-axis at exactly one point (one x-intercept)
2. INFER the mathematical condition needed
- For a quadratic to have exactly one x-intercept, it must have exactly one real solution
- This happens when the discriminant equals zero: \(\Delta = 0\)
- We need to use the discriminant formula: \(\Delta = \mathrm{b^2 - 4ac}\)
3. TRANSLATE the coefficients from our function
- From \(\mathrm{f(x) = -8x^2 + 16x + k}\), we identify:
- \(\mathrm{a = -8}\)
- \(\mathrm{b = 16}\)
- \(\mathrm{c = k}\)
4. SIMPLIFY by setting up and solving the discriminant equation
- Calculate discriminant: \(\Delta = 16^2 - 4(-8)(k)\)
- \(\Delta = 256 + 32k\)
- Set equal to zero: \(256 + 32k = 0\)
- Solve for k: \(32k = -256\)
- \(\mathrm{k = -8}\)
Answer: -8
Why Students Usually Falter on This Problem
Most Common Error Path:
Weak TRANSLATE skill: Students may not recognize that "touches the ground at exactly one point" means the parabola has exactly one x-intercept. Instead, they might think it means finding where the parabola equals some positive value, or they might try to find the vertex without connecting it to the single-intercept condition. This leads to confusion and random guessing.
Second Most Common Error:
Missing conceptual knowledge about discriminants: Students might recognize they need one x-intercept but not remember that this occurs when \(\Delta = 0\). They may try to set \(\mathrm{f(x) = 0}\) and attempt to solve without using the discriminant condition. This causes them to get stuck with a quadratic containing the unknown k, leading to guessing.
The Bottom Line:
This problem requires connecting a real-world description ("touching the ground at one point") to a specific algebraic condition (discriminant equals zero). Students who struggle with translating geometric language into algebraic constraints will find this challenging.