Question

Using the completing-the-square method, rewrite f(x) = x2 − 6x + 2 in vertex form.

f(x) = (x − 3)2
f(x) = (x − 3)2 + 2
f(x) = (x − 3)2 − 7
f(x) = (x − 3)2 + 9

Answer 1

To rewrite the quadratic function \( f(x) = x^2 - 6x + 2 \) in vertex form, we will complete the square.

1. **Start with the equation:**
\[
f(x) = x^2 - 6x + 2
\]

2. **Group the \( x \) terms:**
\[
f(x) = (x^2 - 6x) + 2
\]

3. **Complete the square:**
- Take the coefficient of \( x \) (which is -6), divide it by 2 to get -3, and then square it to get 9.
- We will add and subtract 9 inside the parentheses:
\[
f(x) = (x^2 - 6x + 9 - 9) + 2
\]

4. **Rewrite the equation:**
\[
f(x) = ((x - 3)^2 - 9) + 2
\]

5. **Combine the constants:**
\[
f(x) = (x - 3)^2 - 9 + 2
\]
\[
f(x) = (x - 3)^2 - 7
\]

Now we have rewritten the function in vertex form. The final result is:
\[
f(x) = (x - 3)^2 - 7
\]

Thus, the correct option is:
\[
\text{f(x) = (x − 3)² − 7}
\]