Precalculus, Chapter 9, 9.4, Section 9.4, Problem 71

You need to remember what a quadratic model is, such that:
a_n = f(n) = a*n^2 + b*n + c
The problem provides the following information, such that:
a_0 = -3 => f(0) = a*0^2 + b*0 + c => c = -3
a_2 = 1 => f(2) = a*2^2 + b*2 + c => 4a + 2b + c =1
a_4 = 9 => f(4) = a*4^2 + b*4 + c => 16a + 4b + c = 9
You need to replace -3 for c in equation 4a + 2b + c =1:
4a + 2b - 3 = 1 => 4a + 2b = 4 => 2a + b = 2
You need to replace -3 for c in equation 16a + 4b + c = 9 :
16a + 4b - 3 = 9 => 16a + 4b = 12 => 4a + b = 3
Subtract 2a + b = 2 from 4a + b = 3, such that:
4a + b - 2a - b = 3 - 2
2a = 1=> a = 1/2
Replace 1/2 for a in equation 2a + b = 2 such that:
2*(1/2) + b = 2 => 1 + b = 2 => b = 1
Hence, the quadratic model for the given sequence is a_n = (1/2)n^2+ n - 3.