To rewrite the quadratic function $y = 2(x - 6)^2 - 6$ in standard form, we need to expand it. The standard form of a quadratic function is:

$y = ax^2 + bx + c$

Step 1: Expand $(x - 6)^2$

$(x - 6)^2 = x^2 - 12x + 36$

Step 2: Distribute the 2

$2(x^2 - 12x + 36) = 2x^2 - 24x + 72$

Step 3: Subtract 6

$y = 2x^2 - 24x + 72 - 6$

$y = 2x^2 - 24x + 66$

Final Answer:

The quadratic function in standard form is:

$y = 2x^2 - 24x + 66$

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Would you like me to explain the process in more detail or clarify anything?

Related Questions:

1. What is the vertex form of a quadratic function, and how is it related to standard form?
2. How do you find the vertex of the quadratic function in standard form?
3. What does the coefficient of $x^2$ in standard form represent graphically?
4. Can this function be factored easily? If so, how?
5. What are the roots of this quadratic function, and how are they calculated?

Tip: Always simplify constants at the end of the expansion to ensure your function is in the cleanest standard form.