Question

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

Answer 1

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

1. Start with the quadratic expression: \( x^2 - 16x \).
2. Take half of the coefficient of \( x \) (which is -16), square it, and add and subtract this value inside the equation.

Half of -16 is -8, and squaring it gives:
\[
(-8)^2 = 64
\]

3. Rewrite the quadratic expression by adding and subtracting 64:
\[
y = (x^2 - 16x + 64) + 14 - 64
\]

4. Now, this can be factored:
\[
y = (x - 8)^2 - 50
\]

Now we have the equation in vertex form \( y = (x - h)^2 + k \), where \( (h, k) \) is the vertex.

From \( y = (x - 8)^2 - 50 \):
- \( h = 8 \)
- \( k = -50 \)

Thus, the vertex of the quadratic function \( y = x^2 - 16x + 14 \) is at the point \((8, -50)\).