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What is the vertex of the quadratic function y = 2x^2 - 8x + 5?

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Question: What is the vertex of the quadratic function y = 2x^2 - 8x + 5?

Options:

  1. (2, -3)
  2. (4, -3)
  3. (2, 5)
  4. (4, 5)

Correct Answer: (2, -3)

Solution:

The vertex can be found using the formula x = -b/(2a). Here, a = 2 and b = -8, so x = 8/4 = 2. Plugging x = 2 into the equation gives y = 2(2)^2 - 8(2) + 5 = -3. Thus, the vertex is (2, -3).

What is the vertex of the quadratic function y = 2x^2 - 8x + 5?

Practice Questions

Q1
What is the vertex of the quadratic function y = 2x^2 - 8x + 5?
  1. (2, -3)
  2. (4, -3)
  3. (2, 5)
  4. (4, 5)

Questions & Step-by-Step Solutions

What is the vertex of the quadratic function y = 2x^2 - 8x + 5?
  • Step 1: Identify the coefficients a and b from the quadratic function y = 2x^2 - 8x + 5. Here, a = 2 and b = -8.
  • Step 2: Use the formula to find the x-coordinate of the vertex: x = -b / (2a).
  • Step 3: Substitute the values of a and b into the formula: x = -(-8) / (2 * 2).
  • Step 4: Simplify the expression: x = 8 / 4, which equals 2.
  • Step 5: Now, find the y-coordinate of the vertex by plugging x = 2 back into the original equation: y = 2(2)^2 - 8(2) + 5.
  • Step 6: Calculate y: y = 2(4) - 16 + 5, which simplifies to y = 8 - 16 + 5.
  • Step 7: Further simplify: y = -8 + 5, which equals -3.
  • Step 8: Combine the x and y coordinates to find the vertex: The vertex is (2, -3).
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