Question

the equation of a parabola is $y = x^{2} - 6x + 1$. write the equation in vertex form.
write any numbers as integers or simplified proper or improper fractions.

Explanation:

Step1: Recall vertex form

The vertex form of a parabola is \( y = a(x - h)^2 + k \), where \((h, k)\) is the vertex. We need to complete the square for the given quadratic equation \( y = x^2 - 6x + 1 \).

Step2: Group \(x\)-terms

Group the \(x\)-terms: \( y=(x^2 - 6x)+1 \)

Step3: Complete the square

To complete the square for \( x^2 - 6x \), take half of the coefficient of \(x\) (which is \(-6\)), square it. Half of \(-6\) is \(-3\), and \((-3)^2 = 9\). Add and subtract 9 inside the parentheses:
\( y=(x^2 - 6x + 9 - 9)+1 \)

Step4: Rewrite as perfect square

Rewrite \( x^2 - 6x + 9 \) as \((x - 3)^2\):
\( y=((x - 3)^2 - 9)+1 \)

Step5: Simplify

Simplify the constants: \( y=(x - 3)^2 - 9 + 1=(x - 3)^2 - 8 \)

Answer:

\( y=(x - 3)^2 - 8 \)