A quadratic function in standard fomr is:
[tex]y=ax^2+bx+c[/tex]
a is the number multiplying the quadratic term. b is the number multiplying the x and c is the term without x.
In this case we have:
[tex]f(x)=-x+6-x^2[/tex]
Now if we rearrange as above:
[tex]f(x)=-x^2-x+6[/tex]
And now we can clearly see that:
a = -1
b = -1
c = 6
now for the table we need to input each term in it. The third row in the x section of the table will be always 6, because is a constant which does not depend of x.
We need to use additional values to the given ones to complete the interval [-4, 3].
We'll use:
-4, -3, -2, -1, -0.5, 0, 1, 2
Now, for -x², the row is:
[tex]\begin{gathered} -(-4)^2=-16 \\ -(-3)^2=-9 \\ -(-2)^2=-4 \\ -(-1)^2=-1 \\ -(-0.5)^2=-0.25 \\ -(0)^2=0 \\ -(1)^2=-1 \\ -(2)^2=-4 \\ -(3)^2=-9 \end{gathered}[/tex]
The -x row is:
[tex]\begin{gathered} -(-4)=4 \\ -(-3)=3 \\ -(-2)=2 \\ -(-1)=1 \\ -(-0.5)=0.5 \\ -(0)=0 \\ -(1)=-1 \\ -(2)=-2 \\ -(3)=-3 \end{gathered}[/tex]
Finally we need to complete the y row. This is the sum of the three values in the x:
The y row is:
[tex]\begin{gathered} -16+4+6=-6 \\ -9+3+6=0 \\ -4+2+6=4 \\ -1+1+6=6 \\ -0.25+0.5+6=6.25 \\ 0+0+6=6 \\ -1-1+6=4 \\ -4-2+6=0 \\ -9-3+6=6 \end{gathered}[/tex]
With that, the table is complete, and the problem is solved.
