Question

Algebra 2 help! Medal

Answer 1

@ganeshie8

Answer 2

where is f(x) ?

Answer 3

@atoxicguy

Answer 4

I think it is the same as the last problem lemme check

Answer 5

yes it is the same: x^2+6x-16

Answer 6

\(\large f(x) = x^2 + 6x - 16\)

Answer 7

our goal is to convert it to vertex form :
\(\large f(x) = (x-h)^2 + k\)

Answer 8

take half of x coefficient,
square it, then add and subtract

Answer 9

\(\large f(x) = x^2 + 6x - 16\)

half of x coefficient = 6/2 = 3
square = 3^2

\(\large f(x) = x^2 + 6x + 3^2 - 3^2- 16\)

Answer 10

Next use the identity : \(a^2 + 2ab + b^2 = (a+b)^2\)

Answer 11

\(\large f(x) = x^2 + 6x + 3^2 - 3^2- 16\)

\(\large f(x) = (x+3)^2 - 3^2- 16\)

Answer 12

ok I get it so far

Answer 13

then what?

Answer 14

how do we get the (x-h)^2+k ?

Answer 15

\(\large f(x) =(x+3)^2 - 3^2- 16\)

\(\large f(x) =(x+3)^2 - 25\)

\(\large f(x) =(x--3)^2 -25\)

Answer 16

h = -3
k = -25

Answer 17

so f(x)=(x+3)^2-25?

Answer 18

Yes ! thats the vertex form !