Question

Complete the square to rewrite y = x2 – 6x + 5 in vertex form. Then state whether the vertex is a maximum or a minimum and give its coordinates. A. Maximum at (3, –4) B. Minimum at (3, –4) C. Minimum at (–3, –4) D. Maximum at (–3, –4)

Answer 1

X2 - 6x + 5

= (x - 3)^2 - 9 + 5

= (x - 3)^2  - 4

vertex is a minimum  at  (3, -4)

(

Answer 2

Answer:

B. Minimum at (3, –4)

Step-by-step explanation:

Given is a quadratic function as

$y=x^2-6x+5$

Considering the first two terms, we find that if we add 9 we can make it a perfect square

Hence add and subtract 9 to right side

$y=x^2-6x+9-9+5\\ =(x-3)^2-4$

This is in vertex form

Vertex = (3,-4)

Since coefficient of leading term of x is positive, the parabola is open up and hence minimum is at (3,-4)

Option B,