Which equation is y = 3x^2 + 22x + 52 rewritten in vertex form?

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 given by: y = a(x – h)² + k, where (h, k) is the vertex of the parabola.

We start with the given equation:

y = 3x² + 22x + 52

First, factor out the coefficient of x² from the first two terms:

y = 3(x² + (22/3)x) + 52

Next, to complete the square for the expression inside the parentheses, we take the coefficient of x, which is 22/3, divide it by 2 (getting 11/3), and then square it (getting 121/9).

Add and subtract 121/9 inside the parentheses:

y = 3(x² + (22/3)x + 121/9 - 121/9) + 52

This simplifies to:

y = 3((x + 11/3)² - 121/9) + 52

Now, distribute the 3:

y = 3(x + 11/3)² - 3(121/9) + 52