Asked by Steve
find a quadratic function in standard form for each set of points.
(-1,2),(0,1),(1,-4)
(-1,2),(0,1),(1,-4)
Answers
Answered by
bobpursley
y=ax^2+bx+c
a) 2=a-b+c
b) 1= c
c) -4=a+b+c
from equation b< c=1
add a)+c)
-2=2a+2*1
a=-2
then in equation c)
-4=-2+b+1
b= -3
y=ax^2+bx + c
y=-2x^2-3x+1
a) 2=a-b+c
b) 1= c
c) -4=a+b+c
from equation b< c=1
add a)+c)
-2=2a+2*1
a=-2
then in equation c)
-4=-2+b+1
b= -3
y=ax^2+bx + c
y=-2x^2-3x+1
Answered by
Steve
thank you very much
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