Question

Enter the values of h and k so that y = x2 + 6x + 10 is in vertex form.

y = (x + ) {blank} 2 + {blank}

Answer 1

y = x^2 + 6x + 10
y = (x + 3)^2 - 9 + 10
y = (x + 3)^2 + 1

Answer 2

we have

`y=x^(2) +6x+10`

we know that

the equation in vertex form is equal to

`y=(x-h)^(2) +k\\`

where

`(h,k)` is the vertex

Group terms that contain the same variable, and move the constant to the opposite side of the equation

`y-10=x^(2) +6x`

Complete the square. Remember to balance the equation by adding the same constants to each side.

`y-10+9=x^(2) +6x+9`

`y-1=x^(2) +6x+9`

Rewrite as perfect squares

`y-1=(x+3)^(2)`

`y=(x+3)^(2)+1`

`(h,k)=(-3,1)`

therefore

the answer is

the equation in vertex form is equal to

`y=(x+3)^(2)+1`