What is a equation for this data. X=0,Y=2,X=1,Y=120,and X=2,Y=7200
A parabola y=ax2+bx+c is a equation. We need to calulate a, b and c. $$ We have x1=0 and y1=2 and calculate y1=ax21+bx1+c⇒2=0+0+c⇒c=2 $$ We also have x2=1 and y2=120 and calculate y2=ax22+bx2+2⇒120=a+b+2 $$ ⇒a+b=118(1) $$ Last but not least x3=2 and y3=7200 and calculate y3=ax23+bx3+2⇒7200=a∗4+b∗2+2 $$ ⇒4a+2b=7198(2) $$(1)a+b=118|∗2(1)2a+2b=236(2)4a+2b=7198(2)−(1):4a−2a=2a=7198−236a=3481a+b=1183481+b=118b=118−3481b=−3363 The equation is \boxed{ y=3481x2−3363x+2 } $$ Proof: y(0)=0−0+2=2 okay! $$ Proof: y(1)=3481−3363+2=120 okay! $$ Proof: y(2)=3481∗4−3363∗2+2=7200 okay!
Each successive integer produces a "y" that is 60 times the previous one...so we have....
[Note...I edited my answer after noticing a small "glitch" in my previous one..!! ]
What is a equation for this data. X=0,Y=2,X=1,Y=120,and X=2,Y=7200
A parabola y=ax2+bx+c is a equation. We need to calulate a, b and c. $$ We have x1=0 and y1=2 and calculate y1=ax21+bx1+c⇒2=0+0+c⇒c=2 $$ We also have x2=1 and y2=120 and calculate y2=ax22+bx2+2⇒120=a+b+2 $$ ⇒a+b=118(1) $$ Last but not least x3=2 and y3=7200 and calculate y3=ax23+bx3+2⇒7200=a∗4+b∗2+2 $$ ⇒4a+2b=7198(2) $$(1)a+b=118|∗2(1)2a+2b=236(2)4a+2b=7198(2)−(1):4a−2a=2a=7198−236a=3481a+b=1183481+b=118b=118−3481b=−3363 The equation is \boxed{ y=3481x2−3363x+2 } $$ Proof: y(0)=0−0+2=2 okay! $$ Proof: y(1)=3481−3363+2=120 okay! $$ Proof: y(2)=3481∗4−3363∗2+2=7200 okay!