prismlearning.academy Logo
NEUR
N

\(\mathrm{f(x) = x^2 - 48x + 2,304}\) What is the minimum value of the given function?...

GMAT Advanced Math : (Adv_Math) Questions

Source: Practice Test
Advanced Math
Nonlinear functions
HARD
...
...
Notes
Post a Query

\(\mathrm{f(x) = x^2 - 48x + 2,304}\)

What is the minimum value of the given function?

Enter your answer here
Solution

1. INFER the function type and behavior

  • Given: \(\mathrm{f(x) = x^2 - 48x + 2,304}\)
  • This is a quadratic function in standard form \(\mathrm{f(x) = ax^2 + bx + c}\)
  • Since \(\mathrm{a = 1 \gt 0}\), the parabola opens upward and has a minimum value

2. INFER the solution strategy

  • For quadratics with \(\mathrm{a \gt 0}\), the minimum occurs at the vertex
  • Complete the square to convert to vertex form \(\mathrm{f(x) = a(x - h)^2 + k}\)
  • The minimum value will be k

3. SIMPLIFY by completing the square

  • Start with: \(\mathrm{f(x) = x^2 - 48x + 2,304}\)
  • To complete the square, take half the coefficient of x: \(\mathrm{-48/2 = -24}\)
  • Square this value: \(\mathrm{(-24)^2 = 576}\)
  • Add and subtract 576 inside the function:
    \(\mathrm{f(x) = x^2 - 48x + 576 + 2,304 - 576}\)
  • Group the perfect square trinomial:
    \(\mathrm{f(x) = (x - 24)^2 + 1,728}\)

4. INFER the minimum value from vertex form

  • The function is now in vertex form: \(\mathrm{f(x) = (x - 24)^2 + 1,728}\)
  • Since \(\mathrm{(x - 24)^2 ≥ 0}\) for all real numbers, the minimum value of this expression is 0
  • This occurs when \(\mathrm{x = 24}\)
  • Therefore, the minimum value is \(\mathrm{0 + 1,728 = 1,728}\)

Answer: 1,728




Why Students Usually Falter on This Problem

Most Common Error Path:

Weak SIMPLIFY skill: Students make arithmetic errors when completing the square. They might incorrectly compute \(\mathrm{(-48/2)^2}\) or make mistakes in the addition/subtraction when rearranging terms. For example, they might miscalculate \(\mathrm{2,304 - 576 = 1,728}\) and get a different constant term, leading to an incorrect minimum value.

Second Most Common Error:

Poor INFER reasoning: Students don't recognize that they need to find the vertex of the parabola to determine the minimum. They might try to set \(\mathrm{f(x) = 0}\) to find x-intercepts instead, or confuse maximum vs minimum. This leads to confusion and guessing rather than systematic solution.

The Bottom Line:

This problem requires both strategic thinking about quadratic behavior and careful execution of completing the square. Students must recognize what "minimum value" means for a quadratic function and then accurately perform multi-step algebraic manipulation.

Rate this Solution
Tell us what you think about this solution
...
...
Forum Discussions
Start a new discussion
Post
Load More
Similar Questions
Finding similar questions...
Previous Attempts
Loading attempts...
Similar Questions
Finding similar questions...
Parallel Question Generator
Create AI-generated questions with similar patterns to master this question type.