Answer
a) Answers may vary, but we will assume the equation is $y=x^2+2x+3$.
b) $(1,6)$, $(2,11)$, $(3,18)$
c) It is assumed you traded the three data points with another team. The solution pairs you get will vary, and we will assume we received the following three points: $(1,5)$, $(2,12)$, $(3,27)$
d) $[4, -5, 6]$
e) We were told by the other group that our equation was correct.
Work Step by Step
b)
$x=1$
$y=x^2+2x+3$
$y=1^2+2*1+3$
$y=1+2+3$
$y=6$
$x=2$
$y=x^2+2x+3$
$y=2^2+2*2+3$
$y=4+4+3$
$y=11$
$x=3$
$y=x^2+2x+3$
$y=3^2+2*3+3$
$y=9+6+3$
$y=18$
d)
$(1,5)$
$y=ax^2+bx+c$
$5=a*1^2+b*1+c$
$5=a*1+b+c$
$5=a+b+c$
$(2,12)$
$y=ax^2+bx+c$
$12=a*2^2+b*2+c$
$12=4a+2b+c$
$(3,27)$
$y=ax^2+bx+c$
$27=a*3^2+b*3+c$
$27=a*9+3b+c$
$27=9a+3b+c$
$a+b+c=5$
$4a+2b+c=12$
$9a+3b+c=27$
$4a+2b+c=12$
$3a+a+b+b+c=12$
$3a+b+(a+b+c)=12$
$3a+b+5=12$
$3a+b+5-5=12-5$
$3a+b=7$
$9a+3b+c=27$
$8a+a+2b+b+c=27$
$8a+2b+(a+b+c)=27$
$8a+2b+5=27$
$8a+2b+5-5=27-5$
$8a+2b=22$
$3a+b=7$
$8a+2b=22$
$3a+b=7$
$3a+b-3a=7-3a$
$b=7-3a$
$8a+2b=22$
$8a+2(7-3a)=22$
$8a+2*7+2*-3a=22$
$8a+14-6a=22$
$2a+14=22$
$2a+14-14=22-14$
$2a=8$
$2a/2=8/2$
$a=4$
$b=7-3a$
$b=7-3*4$
$b=7-12$
$b=-5$
$5=a+b+c$
$5=4-5+c$
$5=-1+c$
$5+1=-1+c+1$
$6=c$
$[4, -5, 6]$
