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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)

User Shawndell
by
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2 Answers

4 votes
x2 - 6x + 5

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

= (x - 3)^2 - 4

vertex is a minimum at (3, -4)

(
User Overmeulen
by
7.0k points
4 votes

Answer:

B. Minimum at (3, –4)

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,

User Youhans
by
6.5k points
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