Question

the equation of a parabola is $y = x^2 - 6x + 4$. write the equation in vertex form.
write any numbers as integers or simplified proper or improper fractions.

Explanation:

Step1: Recall vertex form and complete the square

The vertex form of a parabola is \( y = a(x - h)^2 + k \), where \((h,k)\) is the vertex. For the given equation \( y = x^2 - 6x + 4 \), we complete the square for the \( x \)-terms. First, take the coefficient of \( x \), which is \(-6\), divide by 2: \(\frac{-6}{2}=-3\), and then square it: \((-3)^2 = 9\).

Step2: Rewrite the equation by adding and subtracting the square

We can rewrite the equation as \( y=(x^2 - 6x+9)-9 + 4\). The expression \( x^2 - 6x + 9\) is a perfect square trinomial, which factors to \((x - 3)^2\). Then simplify the constants: \(-9 + 4=-5\). So the equation becomes \( y=(x - 3)^2-5\).

Answer:

\( y=(x - 3)^2-5 \)