Find the local minima of f(x) = x^2 - 4x + 5.

Practice Questions

Q1
Find the local minima of f(x) = x^2 - 4x + 5.
  1. (2, 1)
  2. (1, 2)
  3. (0, 5)
  4. (4, 0)

Questions & Step-by-Step Solutions

Find the local minima of f(x) = x^2 - 4x + 5.
  • Step 1: Identify the function we are working with, which is f(x) = x^2 - 4x + 5.
  • Step 2: Recognize that this is a quadratic function in the form of ax^2 + bx + c.
  • Step 3: Determine the coefficients: a = 1, b = -4, and c = 5.
  • Step 4: Use the formula for the x-coordinate of the vertex, which is x = -b/(2a).
  • Step 5: Substitute the values of a and b into the formula: x = -(-4)/(2*1) = 4/2 = 2.
  • Step 6: Now that we have the x-coordinate of the vertex (x = 2), we need to find the y-coordinate by substituting x back into the function.
  • Step 7: Calculate f(2) = 2^2 - 4*2 + 5 = 4 - 8 + 5 = 1.
  • Step 8: The vertex of the function is at the point (2, 1), which is the local minimum.
  • Quadratic Functions – Understanding the properties of quadratic functions, including how to find their vertex and determine local minima or maxima.
  • Vertex Formula – Using the vertex formula x = -b/(2a) to find the x-coordinate of the vertex for a quadratic function.
  • Function Evaluation – Evaluating the function at the vertex to find the corresponding y-coordinate.
Soulshift Feedback ×

On a scale of 0–10, how likely are you to recommend The Soulshift Academy?

Not likely Very likely