How would I write quadratic fx: 3x^2-18x+25 in vertex form? Please list steps.

1. Start with the quadratic function:

f(x) = 3x^2 - 18x + 25

2. Factor out the coefficient of the \(x^2\) term from the first two terms:

f(x) = 3(x^2 - 6x) + 25

3. Complete the square inside the parentheses:

- Take half of the linear term (coefficient of \(x\)), square it, and add/subtract that value inside the parentheses.

- The linear term is -6, so half of it is -3, and squaring it yields 9.

f(x) = 3(x^2 - 6x + 9 - 9) + 25

4. Rewrite the equation to reflect the perfect square trinomial and simplify:

The perfect square trinomial is \((x - 3)^2\), so:

f(x) = 3((x - 3)^2 - 9) + 25

5. Distribute the 3 and combine constants:

f(x) = 3(x - 3)^2 - 27 + 25

f(x) = 3(x - 3)^2 - 2

6. The vertex form of the quadratic function is:

f(x) = 3(x - 3)^2 - 2

7. Answer: f(x) = 3(x - 3)^2 - 2

In this form, the vertex (h, k) is (3, -2)