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

Answer 1

To identify the vertex of the quadratic equation \( y = x^2 - 16x + 14 \) by completing the square, we follow these steps:

1. Start with the equation:

   \[ y = x^2 - 16x + 14 \]

2. Focus on the quadratic and linear terms, \( x^2 - 16x \). To complete the square, take half of the coefficient of \( x \) (which is \(-16\)), square it, and add and subtract that value:

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

3. Rewrite the equation, adding and subtracting \( 64 \):

   \[ y = (x^2 - 16x + 64) - 64 + 14 \]

4. Simplify the equation:

   \[ y = (x - 8)^2 - 64 + 14 \] \[ y = (x - 8)^2 - 50 \]

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

From our completed square form \( y = (x - 8)^2 - 50 \), we can identify:

• The vertex \((h, k)\) is \((8, -50)\).

Thus, the vertex of the parabola given by the equation \( y = x^2 - 16x + 14 \) is \(\boxed{(8, -50)}\).