Please be detailed
2006-08-12 15:11:02 · 6 answers · asked by franciscoeleodoro 1 in Science & Mathematics ➔ Mathematics
DO YOUR OWN HOMEWORK!
2006-08-12 15:15:47 · answer #1 · answered by Anonymous · 0⤊ 0⤋
assuming f(x)= -2x^2 + 6x+2
f(x)= -2{x^2 - 3x -1 }
= -2{ [ x - (3/2) ] ^2 -1 - (9/4) } half the coefficient of x is - (3/2)
and it's square is (9/4)
f(x) = -2 { [ x- (3/2) ] ^2 -1 - (9/4) }
= -2 { [ x- (3/2) ] ^2 - (13/4) }
= -2 { [ x- (3/2) ] ^2} + (13/2)
2006-08-13 00:08:10 · answer #2 · answered by mth2006to 3 · 0⤊ 0⤋
f(x) = -2x^2 + 6x + 2
f(x) = (-2x^2 + 6x) + 2
f(x) = -2(x^2 - 3x) + 2
f(x) = -2(x^2 - 3x + (9/4) - (9/4)) + 2
f(x) = -2(x^2 - 3x + (9/4)) + (9/2) + 2
f(x) = -2(x - (3/2))^2 + (13/2)
2006-08-12 22:38:30 · answer #3 · answered by Sherman81 6 · 0⤊ 0⤋
There seems to be a problem in your question.
2006-08-12 22:24:14 · answer #4 · answered by Arkangyle 4 · 0⤊ 0⤋
That makes no sense.
2006-08-12 22:16:28 · answer #5 · answered by shmux 6 · 1⤊ 0⤋
f(x)= -2(x^ -3x -1) = -2[(x -1.5)^ - 3.25] = -2(x - 1.5)^ + 6.5
2006-08-12 22:23:54 · answer #6 · answered by grsym 2 · 0⤊ 0⤋