Since the value of x are odd and are equally spaced we change

X to x by the relation x=X-5 and put Y=y

Let the equation of parabola be

$y=a + bx + cx^2.$ Then the normal equation are

$\displaystyle \sum y_i = Na + b \sum xi + c \sum x^2_i$

$\displaystyle \sum xi\ yi=a \sum xi + b \sum x^2_i + c \sum x^3_i$

$\displaystyle \sum xi\ yi =a \sum x_i + b \sum x^2_i + c \sum x^3_i$

| $X$ | $x_i$ | $y_i$ | $x^2_i$ | $x^3_i$ | $x^4_i$ | $x_iy_i$ | $x^2_iy_i$ | | --- | --- | --- | --- | --- | --- | --- | --- | | 1 | -4 | 2 | 16 | -64 | 256 | -8 | 32 | | 2 | -3 | 6 | 9 | -27 | 81 | -18 | 54 | | 3 | -2 | 7 | 4 | -8 | 16 | -14 | 28 | | 4 | -1 | 8 | 1 | -1 | 1 | -8 | 8 | | 5 | 0 | 10 | 0 | 0 | 0 | 0 | 0 | | 6 | 1 | 11 | 1 | 1 | 1 | 11 | 11 | | 7 | 2 | 11 | 4 | 8 | 16 | 22 | 44 | | 8 | 3 | 10 | 9 | 27 | 81 | 30 | 90 | | 9 | 4 | 9 | 16 | 64 | 256 | 36 | 144 | | | 0 | 74 | 60 | 0 | 708 | 51 | 411 | Now the normal equation becomes, $74 = 9a+60c \ \cdots (1)$ $51 = 60b; $ hence $b=0.85 \cdots (2)$ $411 = 60a + 708 c \ \cdots \ (3)$ Now to find a and c multiply (1) by 60 and (3) by 9 $4440=540 a + 3600c$ $3699 = 540a + 6372 c$ $c= - 0.2673$ Now, $9a=74-60c=74+16.038 = 90.038$ Hence $a=10.0042$ The equation of parabola be $y=10.0042 + 0.85 x - 0.2673 x^3$ i.e. $Y=10+0.85 (X-5) - 0.27 x (X-5)^2$ $Y=1+3.55 X - 0.27 X^2$