Question

the equation of a parabola is $y = x^{2} - 6x + 15$. 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 quadratic \( y = x^2 - 6x + 15 \), we complete the square for the \( x \)-terms. First, take the coefficient of \( x \), which is \(-6\), divide by 2: \(\frac{-6}{2}=-3\), then square it: \((-3)^2 = 9\).

Step2: Rewrite the equation by adding and subtracting the square

We can rewrite \( y = x^2 - 6x + 15 \) as \( y=(x^2 - 6x + 9)-9 + 15 \). The \( x^2 - 6x + 9 \) is a perfect square trinomial, which factors to \((x - 3)^2\).

Step3: Simplify the constant terms

Now simplify \(-9 + 15\), which equals \( 6 \). So the equation becomes \( y=(x - 3)^2+6 \).

Answer:

\( y = (x - 3)^2 + 6 \)