Asked by Sade
Find a quadratic model for the sequence.
-4, -4, -3, -1, 2
A.y = 0.5x^2 - 0.5x - 4
B.y = 0.5x^2 - 1.5x - 3
C.y = 4.5x^2 - 21.5x+21
D.y = -4.5x^2 + 21.4x - 21
-4, -4, -3, -1, 2
A.y = 0.5x^2 - 0.5x - 4
B.y = 0.5x^2 - 1.5x - 3
C.y = 4.5x^2 - 21.5x+21
D.y = -4.5x^2 + 21.4x - 21
Answers
Answered by
Reiny
Consider it as the points (1,-4), (2,-4), (3,-3) etc
let the function be ax^2 + bx + c = y
for:
(1,-4) ---> a + b + c = -4 *
(2,-4) ---> 4a + 2b + c = -4 **
(3,-3) ---> 9a + 3b + c = -3 ***
subtract * from **
3a + b = 0 #
subtract ** from ***
5a + b = 1 ##
subtract # from ##
2a = 1
a = 1/2
sub into #
3/2 + b = 0
b = -3/2
sub those two values into * to get
c = -3
so y = (1/2)x^2 - (3/2)x - 3
or
y = .5x^2 - 1.5x - 3 <------ B
or
y = (1/2)(x^2 - 3x - 6)
test it for the last two value of your data
let the function be ax^2 + bx + c = y
for:
(1,-4) ---> a + b + c = -4 *
(2,-4) ---> 4a + 2b + c = -4 **
(3,-3) ---> 9a + 3b + c = -3 ***
subtract * from **
3a + b = 0 #
subtract ** from ***
5a + b = 1 ##
subtract # from ##
2a = 1
a = 1/2
sub into #
3/2 + b = 0
b = -3/2
sub those two values into * to get
c = -3
so y = (1/2)x^2 - (3/2)x - 3
or
y = .5x^2 - 1.5x - 3 <------ B
or
y = (1/2)(x^2 - 3x - 6)
test it for the last two value of your data
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