what is the vertex of the function f(x)=2x^2+4x-16? Write the anser in the form (x,y)

Question

what is the vertex of the function f(x)=2x^2+4x-16?  Write the anser in the form (x,y)

.sam. · Answer

Complete the square

.sam. · Answer

or use -b/2a

anonymous · Answer

i don't get it

.sam. · Answer

2x^2+4x-16
2(x^2+2x)-16
2(x+1)^2-1(2)-16
2(x+1)^2-18

vertex at (-1,-18)

anonymous · Answer

thank you....i'm trying to get what you did.  You added 1. Why?

anonymous · Answer

ok...i see. i had

.sam. · Answer

Its completing the square , did you learn it?

anonymous · Answer

the teacher went over it but i just couldn't get it!!  I think I have it a little better now. at least i hope before my final.

.sam. · Answer

The technique goes by

ax^2+bx+c=0
The coefficient of x^2 must be 1, so factor 'a' out

a(x^2+bx/a)+c=0
Then here's the technique, divide bx/a by 2,then set the x^2 becomes x only,eliminate the x from b/a, then square the whole bracket 

you ended up with 

a(x+b/2a)^2+c=0

Still haven't finished, you need to bring the b/2a OUT the brackets and square it, and when you bring it out, it is always negative, be sure that the 'a' will be multiplying the one you bought it out

a(x+b/2a)^2-(b/2a)^2(a)+c=0