Question

Please solve and find the equation and put the equation in quadratic form

Answer 1

To find the solution to the question we would use the general representation of a quadratic equation below:

`ax^2+bx+c=y`

We then pick a suitable point

When x=0 and y =16

`\begin{gathered} a(0)^2+b(0)+c=16 \\ c=16 \end{gathered}`

We then pick two more suitable points, then solve the resulting simultaneous equation

When x= 3 and y= 0

we have

`\begin{gathered} a(3)^2+b(3)+16=0 \\ 9a+3b=-16 \end{gathered}`

Also, when x=-5 and y =0

we have

`\begin{gathered} a(-5)^2+b(-5)+16=0 \\ 25a-5b=-16 \\ \end{gathered}`

We have gotten two equations, we will solve them simultaneously using the elimination method.

We will multiply equation one by 5 and equation two by 3

`\begin{gathered} 5(9a+3b=-16)_{} \\ 3(25a-5b=-16) \end{gathered}`
`\begin{gathered} 45a+15b=-80 \\ 75a-15b=-48 \\ Add\text{ equation 2 from 1} \\ 120a=-128 \\ a=-(128)/(120) \\ a=(-16)/(15) \end{gathered}`

Substitute the value of a in equation 1

`\begin{gathered} 45(-(16)/(15))+15b=-80 \\ 3*(-16)+15b=-80 \\ -48+15b=-80 \\ 15b=-80+48 \\ 15b=-32 \\ b=-(32)/(15) \end{gathered}`

We will then substitute a and b and c in the representation of the quadratic equation.

`\begin{gathered} -(16)/(15)x^2-(32)/(15)x+16=0 \\ Multiply\text{ through by 15} \\ -16x^2-32x+240=0 \\ therefore\text{ we have} \\ -x^2-2x+15=0 \end{gathered}`