Give an equation of the curve which having the points (1,3), (2,5) and (3,11)?

Give an equation Standard form is y=a*b^x+c of the curve which having the points (1，3)
， (2，5) and (3，11)?
Sol
y=a*b^x+c
3=a*b^1+c
5=a*b^2+c
11=a*b^3+c
(a*b^2+c)－(a*b^1+c)=5－3
a*(b^2－b)=2
(a*b^3+c)－(a*b^1+c)=11－3
a*(b^3－b)=8
(b^3－b)/(b^2－b)=8/2
b^3－b=4b^2－4b
b^3－4b^2+3b=0
b(b^2－4b+3)=0
b(b－1)(b-3)=0
b=0 or b=1 or b=3
(1) b=0
y=c(不合)
(2) b=1
y=a+c(不合)
(3) b=3
a*(27－3)=8
a=1/3
c=3－a*b=2
y=(1/3)*3^x+2

or
題目改為
二次曲線y=ax^2+bx+c通過(1，3) (2，5) and (3，11)，求此二次曲線
Sol
y=a(x－1)(x－2)+p(x－1)+3
y(2)=p*(2－1)+3=5
p=2
y=a(x－1)(x－2)+2x+1
y(3)=a*2*1+6+1=11
a=2
y=2(x－1)(x－2)+2x+1
=2(x^2－3x+2)+2x+1
=2x^2－4x+5

y(x) = a*x^2 + b*x + c


y(1) = a + b + c = 3

y(2) = 4a + 2b + c = 5

y(3) = 9a + 3b + c = 11



y(2) - y(1) = 3a + b = 2

y(3) - y(2) = 5a + b = 6


Subtract: a = 2

=> b = 2 - 6 = -4

=> c = 3 + 4 - 2 = 5

=> y(x) = 2x^2 - 4x + 5 = Answer