Question

Complete the square to rewrite y = x^2 – 6x + 14 in vertex form. Then state whether the vertex is a maximum or minimum and give its coordinates.

Answer 1

y=x²-6x+14
y=(x²-2*3x +3²-3²)+14
y=(x²-2*3x+3²)+14-9
y=(x-3)²+5

a=1
p=3
q=5


a>0 ⇒ vertex is minimum

V(p,q) ⇒ V(3;5)




Regards M.Y.

Answer 2

Answer:

Vertex is minimum.

Coordinate is (3,5)

Step-by-step explanation:

Given :  $y = x^2-6x + 14$

To find : Complete the square to rewrite in vertex form. Then state whether the vertex is a maximum or minimum and give its coordinates.

Solution :

The general vertex form is $y=a(x-h)^2+k$

where (h,k) is the vertex of the function

Converting into vertex form by completing the square,

$y = x^2-6x + 14$

$y = (x^2-2\times 3x+3^2-3^2)+14$

$y = (x^2-2\times 3x+3^2)+14-9$

$y = (x-3)^2+5$

This is the vertex form where a=1, (h,k)=(3,5)

If a>0 then vertex is minimum

If a<0 then vertex is maximum.

In our case, 1>0 then vertex is minimum.

Coordinates is (3,5).