Algebra 2. Please help me.

Quote:

what is the vertex of the function: f(x)=2x^2+4x-16

the quick and dirty way of finding the vertex of ax^2 + bx + c is that it lies at (-b/2a, -b^2/(4a) + c), in this case (-1, -18)

that being said learning how to complete the square is worth your time and effort

f(x) = 2x^2 + 4x - 16

first factor the coefficient of the x^2 term out front to get

f(x) = 2(x^2 + 2x - 8)

now we want to put what's in the parentheses into (x-a)^2 + b form. we note that this is (x^2 + 2ax + a^2) + b and in this case

f(x) = 2( (x+1)^2 - 1 -8) = 2( (x+1)^2 - 9) = 2(x+1)^2 - 18

from this we can immediately read off the vertex as the point where the (x+1)^2 term is 0, i.e. at x=-1, y=-18

Brooklynbleach:

what is the vertex of the function: f(x)=2x^2+4x-16

Umm, Thanks Rom, I haven't see that way of getting y before.

I'll show you how I'd do it.

a=2, b=4, c=16

x = -b/(2a)
x = -4/(2*2) = -1
then sub x=-1 into the function to find y.

I can explain a bit more about why x=-b/(2a) if you want me too.