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Write an equation of the quadratic that passes through the points (1, -6), (2, 3), and (5, -6). Please show all your steps carefully and clearly. Write your final answer in standard form.

A. \(y = -x^2 + 6x - 1\)
B. \(y = x^2 - 6x + 1\)
C. \(y = -x^2 + 6x - 9\)
D. \(y = x^2 - 6x - 9\)

User JD Allen
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1 Answer

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Final answer:

Using the given points to create a system of equations based on the form y = ax^2 + bx + c, we find that the coefficients a = -1, b = 6, and c = -11, giving us the quadratic equation y = -x^2 + 6x - 11. However, this solution is not offered in the provided answer options, suggesting an error in the question or options provided.

Step-by-step explanation:

To find the equation of a quadratic that passes through the points (1, -6), (2, 3), and (5, -6), we assume a general form of y = ax^2 + bx + c. We will substitute the given points into this equation to get a system of equations to solve for the coefficients a, b, and c.

  • Substituting (1, -6): -6 = a(1)^2 + b(1) + c
  • Substituting (2, 3): 3 = a(2)^2 + b(2) + c
  • Substituting (5, -6): -6 = a(5)^2 + b(5) + c

Solving the system of equations, we get:

  1. -6 = a + b + c
  2. 3 = 4a + 2b + c
  3. -6 = 25a + 5b + c

This can be solved using methods such as substitution or elimination. Upon solving, we find that a = -1, b = 6, and c = -11. Therefore, the quadratic equation is y = -x^2 + 6x - 11, which is not listed in the provided options, indicating a possible mistake in the question or the options.

User Superoryco
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9.2k points