Question

the equation of a parabola is $y = x^2 - 8x + 16$. 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 \( y = x^2 - 8x + 16 \).

Step2: Complete the square (or recognize perfect square)

Notice that \( x^2 - 8x + 16 \) is a perfect square trinomial. Recall that \( (a - b)^2 = a^2 - 2ab + b^2 \). Here, \( a = x \), \( 2ab = 8x \), so \( 2b = 8 \) implies \( b = 4 \). Then \( b^2 = 16 \), which matches the constant term. So \( x^2 - 8x + 16=(x - 4)^2 \).

Step3: Write in vertex form

Substituting back into the equation, we get \( y=(x - 4)^2+0 \) (since there's no additional constant term after factoring the perfect square).

Answer:

\( y = (x - 4)^2 \)