Complete the square to identify the vertex of y = x^2 - 16x + 14.

Answer 1

To complete the square for the quadratic equation \( y = x^2 - 16x + 14 \), we follow these steps:

1. Identify the coefficients: The equation is in the form \( y = ax^2 + bx + c \), where \( a = 1 \), \( b = -16 \), and \( c = 14 \).

2. Isolate the \( x \) terms: We focus on the \( x^2 - 16x \) part.

3. Complete the square: To complete the square for \( x^2 - 16x \):

   • Take half of the coefficient of \( x \) (which is \(-16\)), yielding \(-8\).
   • Square that number: \( (-8)^2 = 64 \).

   Now we rewrite \( x^2 - 16x \) as: \[ x^2 - 16x = (x - 8)^2 - 64 \]

4. Rewrite the original equation: Substitute back into the equation: \[ y = (x - 8)^2 - 64 + 14 \]

5. Combine constant terms: \(-64 + 14 = -50\), so: \[ y = (x - 8)^2 - 50 \]

6. Identify the vertex: The equation is now in vertex form \( y = a(x - h)^2 + k \), where \( (h, k) \) is the vertex. Here, \( h = 8 \) and \( k = -50 \).

Thus, the vertex of the parabola is at the point: \[ \boxed{(8, -50)} \]