Explanation:
The quadratic equation is of the form,
[tex]\begin{gathered} y=ax^2+bx+c \\ a,b,c\text{ are constant} \end{gathered}[/tex]a) For the first option, The given data represents the relation,
[tex]y=-4x[/tex]So, it is not the quadratic equation.
b) From the table, the relation is,
[tex]y=3x[/tex]Not a quadratic equation.
c) to find the quadratic equation,
[tex]\begin{gathered} \text{For (1,2)} \\ 2=a+b+c\ldots\ldots\ldots(1) \\ \text{For (4,8)} \\ 8=a(4)^2+b(4)+c \\ 8=16a+4b+c\ldots\ldots\ldots\text{.}(2) \\ \text{For (7,17)} \\ 17=a(7)^2+b(7)+c \\ 17=49a+7b+c\ldots\ldots\ldots\text{.}(3) \end{gathered}[/tex]Solving these questions,
[tex]\begin{gathered} \text{equation}(1)\times16-equation\text{ (2)} \\ 16a+16b+16c-16a-4b-c=32-8 \\ 12b+15c=24 \\ 4b+5c=8\ldots\ldots\ldots(4) \\ \text{equation (1)}\times49\text{-}equation\text{ (3)} \\ 49a+49b+49c-49a-7b-c=98-17 \\ 42b+48c=81 \\ 14b+16c=27\ldots\ldots(5) \end{gathered}[/tex]Now solve equation (4) and (5),
[tex]\begin{gathered} 14b+16c=27 \\ b=\frac{27-16c}{14} \\ It\text{ gives} \\ 4b+5c=8 \\ 4(\frac{27-16c}{14})+5c=8 \\ 3c+54=56 \\ c=\frac{2}{3} \\ \Rightarrow4b+5c=8 \\ 4b+5(\frac{2}{3})=8 \\ b=\frac{7}{6} \\ \Rightarrow2=a+b+c \\ 2=a+\frac{7}{6}+\frac{2}{3} \\ a=\frac{1}{6} \end{gathered}[/tex]So, the equation is,
[tex]y=\frac{1}{6}x^2+\frac{7}{6}x+\frac{2}{3}[/tex]d) The data represents the relation,
[tex]y=x^3[/tex]Not a quadratic equation.
Answer: option c)
