How can the equation y = 3x^2 + 22x + 52 be rewritten in vertex form?

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To rewrite the quadratic equation y = 3x² + 22x + 52 in vertex form, we need to complete the square. The vertex form of a quadratic equation is expressed as y = a(x – h)² + k, where (h, k) is the vertex of the parabola.

Here’s a step-by-step process on how to complete the square for the given equation:

  1. First, factor out the coefficient of x² from the first two terms: 
    2. y = 3(x² + (22/3)x) + 52
  2. Next, to complete the square, take half of the coefficient of x (which is 22/3), square it, and add and subtract this value inside the parentheses.
    3. Half of (22/3) is (11/3), and squaring it gives (121/9).
  3. Add and subtract (121/9): 
    4. y = 3(x² + (22/3)x + (121/9) - (121/9)) + 52
  4. Now, simplify the equation:
    5. y = 3((x + (11/3))² - (121/9)) + 52
  5. Distribute the 3:
    6. y = 3(x + (11/3))² - 3 * (121/9) + 52
  6. Calculate -3 * (121/9): 
    7. -3 * (121/9) = -363/9 = -40.33 (approximately)
  7. Now, combine it with 52: 
    8. 52 = 468/9.
    So, -40.33 + 52 = 468/9 - 363/9 = 105/9.
  8. Finally, putting all this together, we have:
    9. y = 3(x + (11/3))² + (105/9).

Thus, the equation rewritten in vertex form is:

y = 3(x + (11/3))² + (105/9).

This form clearly shows the vertex of the parabola, which is located at the point (-11/3, 105/9).

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