Question

dentify the vertex for the graph of y = -2x^2 - 12x + 1.
Answer
(-3, 19)
(-3, 55)
(3, -53)
(3, -47)

Answer 1

dentify?

Answer 2

what?

Answer 3

o identify

Answer 4

\[ax^2-bx+c\] make 'a' 1
-2x^2 - 12x + 1
-2(x^2 +6x)+1
-2(x-3)^2-9(-2)+1
-2(x-3)^3+19

Answer 5

can you find the vertex now?

Answer 6

can u walk me through it?

Answer 7

ax^2+bx+c=0
The coefficient of x^2 must be 1 everytime you do completing the square, so you must factor 'a' out

a(x^2+bx/a)+c=0
then, 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

then, 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

Answer 8

k then what?

Answer 9

-2(x-3)^3+19

(x-3)=0
x=3,y=19

Answer 10

I think there're some problems with @.Sam.'s answer....
y = -2x^2 - 12x + 1.
= -2(x^2+6x) +1
= -2 (x^2 + 6x +9 - 9) +1
= -2 (x+3)^2 + 18+1
= -2 (x+3)^2 + 19

Answer 11

yes

Answer 12

isn it -3, 19?

Answer 13

yes

Answer 14

k thanks

Answer 15

yw