Question

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:
\[
\boxed{(8, -50)}
\]