First, we can notice that all the optins have the same x values: 0, 1, 2 and 3.
These are at equal differences of 1.
A quadratic relationship has the property that the difference of the differences of subsequent points is constant.
So, to check wheter it is a quadratic relationship, we can calculate the differences between f(1) and f(0), f(2) and f(1) and f(3) and f(2):
[tex]\begin{gathered} f(1)-f(0) \\ f(2)-f(1) \\ f(3)-f(2) \end{gathered}[/tex]And then we make the difference between the first and second differences and between the third and second. if they are equal and difference than 0, it is a quadratic relationship.
For the first, we have:
[tex]\begin{gathered} f(1)-f(0)=-8-(-4)=-8+4=-4 \\ f(2)-f(1)=-10-(-8)=-10+8=-2 \\ f(3)-f(2)=-10-(-10)=-10+10=0 \end{gathered}[/tex]Now, the difference of differences:
[tex]\begin{gathered} -2-(-4)=-2+4=2 \\ 0-(-2)=2 \end{gathered}[/tex]They are equal, so the first one is a quadratic relationship.
The second, we have:
[tex]\begin{gathered} f(1)-f(0)=0-(-2)=0+2=2 \\ f(2)-f(1)=2-0=2 \\ f(3)-f(2)=4-2=2 \end{gathered}[/tex]Now, the difference of differences:
[tex]\begin{gathered} 2-2=0 \\ 2-2=0 \end{gathered}[/tex]They are the same but they are equal to zero, so the second is not a quadratic relationship. This will happen when the first differences are constant, becaise this is a linear relationship. Notice that the fifth also have constant differences, so the fifth is not a quadratic relationship.
For the third, we have:
e have:
[tex]\begin{gathered} f(1)-f(0)=-4-4=-8 \\ f(2)-f(1)=-4-(-4)=-4+4=0 \\ f(3)-f(2)=4-(-4)=4+4=8 \end{gathered}[/tex]Now, the difference of differences:
[tex]\begin{gathered} 0-(-8)=0+8=8 \\ 8-0=8 \end{gathered}[/tex]They are equal so, the third is also a quadratic relationship.