Question

How would I rewrite this in vertex form and identify the vertex? y = x^2 + 4x - 6

Answer 1

first divide the coefficient of x (6) by 2 which = 2and write:

(x + 2)^2

Answer 2

when this is expanded the last term = 4 but you want it to be -6 so u subtract 10


and the vertex form becomes

(x + 2)^2 - 10

Answer 3

the minimum value of the function is when (x + 2)^2 = 0 so it is -10
the x-coordinate of the minimum value is x = -2
so the minimum is at (-2,-10)

Answer 4

y = x^2 + 4x - 6

y = (x^2 + 4x) - 6 ; complete the square

y = (x^2 + 4x +4) -4 - 6 ; compact it and combine like terms

y = (x+2)^2 -10

completeing the square would take some geometry to help understand what is going on :)

Answer 5

sorry - in my first post, coefficient of x = 4 = to give (x+2)^2

Answer 6

like this :)