Sure. The first step is to rewrite the given quadratic function, f(x) = 3x² - 6x - 1, in a suitable form, ready for completing the square. We can achieve this by separating the constant term, -1, and dividing the entire equation by the coefficient of x², which is 3. Hence, we get f(x) = x² - 2x - 1/3.
Next, we want to complete the square for the x² and x terms. For that, let's recall the standard formula for a square of binomial (x - h)² = x² - 2hx + h².
When we compare this with the transformed function, the coefficient of the x term in the latter is -2. To have the same form as in the square of the binomial, 2h should be equal to 2, from where we find h=1.
To complete the square, we add h² to the equation. After doing this operation, the equation becomes f(x) = (x² - 2x + h²) - 1/3 - h².
By substituting h=1, the equation becomes f(x) = (x² - 2x + 1) - 4/3 = (x - 1)² - 4/3.
Finally, we have g(x) = 3(x-1)² - 4, after we multiply both sides by 3, since, originally, the coefficient of x² was 3 in the given quadratic function.
As a result, the function 3x² - 6x -1 can be rewritten in the form g(x)=a(x-h)²+k as g(x) = 3(x - 1)² - 4.