Bonus: [2 pts] Demonstrate (using linear systems) why it is not possible to find a quadratic polynomial
which passes through these four points.
To see why this is impossible, lets try to find a function that hits all the points and see what happens!
Lets start with the general quadratic function
f (x) = ax2 + bx + c
Now we need a linear system. Lets plug in the points like we did for part 1.
For (−1, −1):
f (−1) = a(−1)2 + b(−1) + c
−1 = a − b + c
For (0, 1):
f (0) = a(0)2 + b(0) + c
1=c
For (1, 3):
f (1) = a(1)2 + b(1) + c
3=a+b+c
For (4, −1):
f (4) = a(4)2 + b(4) + c
−1 = 16a + 4b + c
Now we have a system of 4 equations. We can perform Gaussian Elimination to find the unique solution of
the system. First, lets create an augmented matrix to represent the system:
1 −1 1 −1
0
0 1 1
1
1 1 3
16 4 1 −1
Next, we need to perform row reduction to simplify the matrix before we solve for the solution.
1 −1
1
−1
1 −1
1
−1
R4 :R4 −16R3 0
:R3 ÷2 0
0
1
1
0
1
1
R3−→
−→
0 2
0
4
0 1
0
2
R3 :R3 −R1
R4 :R4 −20R3
0 20 −15 −49
0 0 −15 −89
1 −1 1 −1
1
R :R +15R2
0 0 1
−−4−−4−−−−→
0 1 0
2
0 0 0 −74
During row reduction, we discover that the system is inconsistent. There are no solutions to the system,
which means no quadratic function exists such that all the points are met.
Please see the next page for question 2.
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