Precalculus, Chapter 9, 9.4, Section 9.4, Problem 73

The given sequence is:
a_1 = 0 , a_2=8 , a_4=30
To determine its quadratic model, apply the formula
f(n) = an^2 + bn + c
where f(n) represents the nth term of the sequence, f(n)=a_n .
So, plug-in the first term of the sequence.
0=a(1)^2 + b(1) + c
0=a+b+c (Let this be EQ1.)
Plug-in too the second term of the sequence.
8=a(2)^2+b(2)+c
8=4a+2b+c (Let this be EQ2.)
And, plug-in the 4th term of the sequence.
30=a(4)^2+b(4)+c
30=16a+4b+c (Let this be EQ3.)
To solve for the values of a, b and c, apply elimination method of system of equations. In this method, a variable or variables should be removed.
Let's eliminate c. To do so, subtract EQ1 from EQ2.
EQ2: 8=4a+2b+c
EQ1: -(0=a+b+c)
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8=3a+b (Let this be EQ4.)
Let's eliminate c again. This time, subtract EQ2 from EQ3.
EQ3: 30=16a+4b+c
EQ2: -(8=4a+2b+c)
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22=12a+2b
And this simplifies to:
22/2=(12a+2b)/2
11=6a+b (Let this be EQ5.)
Then, eliminate b. To do so, subtract EQ4 from EQ5.
EQ5: 11=6a+b
EQ4: -(8=3a+b)
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3=3a
Isolating the a, it becomes:
3/3=(3a)/3
1=a
Then, plug-in the value of a to either EQ4 or EQ5. Let's use EQ4.
8=3a+b
8=3(1) + b
8=3+b
8-3=3-3+b
5=b
And, plug-in the values of a and b to either EQ1, EQ2 or EQ3. Let's use EQ1.
0=a+b+c
0=1+5+c
0=6+c
0-6=6-6+c
-6=c
Now that the values of a, b and c are known, plug-in them to:
f(n)=an^2+bn+c
f(n)=(1)n^2+5n+(-6)
f(n)=n^2+5n-6
Replacing the f(n) with an, it becomes:
a_n=n^2+5n-6
Therefore, the quadratic model of the sequence is a_n=n^2+5n-6 .