Complete the square to identify the vertex of y=x2−16x+14

Question

𝑦
=
𝑥
2
−
16
𝑥
+
14
.(1 point)
Responses

(8,30)
(
8
,
30
)
left parenthesis 8 comma 30 right parenthesis

(8,−50)
(
8
,
−
50
)
left parenthesis 8 comma negative 50 right parenthesis

(8,78)
(
8
,
78
)
left parenthesis 8 comma 78 right parenthesis

(8,14)

Answer 1

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

1. Start with the original equation:
   \[ y = x^2 - 16x + 14 \]

2. Focus on the quadratic and linear terms:
   \[ x^2 - 16x \]

3. To complete the square, take half of the coefficient of \(x\) (which is \(-16\)), square it, and then add and subtract that value inside the equation:
   Half of \(-16\) is \(-8\), and \((-8)^2 = 64\).

4. Rewrite the expression by adding and subtracting 64:
   \[ y = (x^2 - 16x + 64) - 64 + 14 \] \[ y = (x - 8)^2 - 50 \]

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

From the completed square form \(y = (x - 8)^2 - 50\), we can see that the vertex is \((8, -50)\).

So, the vertex of the parabola given by the equation \(y = x^2 - 16x + 14\) is:

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