Hello,
1) we imagine a linear function:
y=ax+b
(1,3)==>3=a+b (1)
(2,6)==>6=2a+b (2)
(2)-(1)==>a=3==>b=3-3=0
So we find y=3x
But (3,11)==>11=33 is false ==> no linear function A,B and D are false.
2) we imagine a quadratic function
y=ax²+bx+c
(1,3)==>3=a+b+c (1)
(2,6)==>6=4a+2b+c (2)
(3,11)==>11=9a+3b+c (3)
(3)-(2)==>5a+b=5 (4)
(2)-(1)==>3a+b=3 (5)
(4)-(5)==> 2a=2==> a=1
==> b=5-51=0
(1)==>3=1+0+c==>c=2
The quadratic function is y=x²+2
and you have make a mistake in your question!
Answer is C if you change in y=x²+2
Proof:
(4,18) 18=14²+04+2=16+2=18
