Question

I need help with this question please. It’s also just a homework practice btw

Answer 1

Step-by-step explanation:

The quadratic equation is of the form,

`\begin{gathered} y=ax^2+bx+c \\ a,b,c\text{ are constant} \end{gathered}`

a) For the first option, The given data represents the relation,

`y=-4x`

So, it is not the quadratic equation.

b) From the table, the relation is,

`y=3x`

Not a quadratic equation.

c) to find the quadratic equation,

`\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}`

Solving these questions,

`\begin{gathered} \text{equation}(1)*16-equation\text{ (2)} \\ 16a+16b+16c-16a-4b-c=32-8 \\ 12b+15c=24 \\ 4b+5c=8\ldots\ldots\ldots(4) \\ \text{equation (1)}*49\text{-}equation\text{ (3)} \\ 49a+49b+49c-49a-7b-c=98-17 \\ 42b+48c=81 \\ 14b+16c=27\ldots\ldots(5) \end{gathered}`

Now solve equation (4) and (5),

`\begin{gathered} 14b+16c=27 \\ b=(27-16c)/(14) \\ It\text{ gives} \\ 4b+5c=8 \\ 4((27-16c)/(14))+5c=8 \\ 3c+54=56 \\ c=(2)/(3) \\ \Rightarrow4b+5c=8 \\ 4b+5((2)/(3))=8 \\ b=(7)/(6) \\ \Rightarrow2=a+b+c \\ 2=a+(7)/(6)+(2)/(3) \\ a=(1)/(6) \end{gathered}`

So, the equation is,

`y=(1)/(6)x^2+(7)/(6)x+(2)/(3)`

d) The data represents the relation,

`y=x^3`

Not a quadratic equation.

Answer: option c)