Question

Find the quadratic model in standard form for the set of values (3, -6), (1,-2), (6,3)

Answer 1

We are asked to determine a quadratic equation given a set of points. To do that, let's remember that a quadratic function is of the following form:

`y=ax^{2^{}}+bx+c`

Now we replace the given points:

Replacing (3, -6)

`\begin{gathered} -6=a(3)^2+3b+c \\ -6=9a+3b+c,(1) \end{gathered}`

Replacing (1, -2)

`\begin{gathered} -2=a(1)^2+b(1)+c \\ -2=a+b+c,(2) \end{gathered}`

Replacing (6, 3)

`\begin{gathered} 3=a(6)^2+b(6)+c \\ 3=36a+6b+c,(3) \end{gathered}`

We get a system of three equations and three variables, these are:

`\begin{gathered} 9a+3b+c=-6,(1) \\ a+b+c=-2,(2) \\ 36a+6b+c=3,(3) \end{gathered}`

Solving for "c" in equation (1):

`c=-6-3b-9a,(4)`

Solving for "c" in equation (2):

`c=-2-b-a,(5)`

Solving for "c" in equation (3)

`c=3-6b-36a,(6)`

equating equations (4) and (5):

`-6-3b-9a=-2-b-a`

Rewritting:

`\begin{gathered} -3b+b-9a+a=-2+6 \\ -8a-2b=4,(7) \end{gathered}`

equating equation (5) and (6):

`-2-b-a=3-6b-36a`

Rewritting:

`\begin{gathered} -b-a+6b+36a=3+2 \\ 35a+5b=5,(8) \end{gathered}`

Now we solve for "a" in equation (7):

`\begin{gathered} -8a=4+2b \\ a=-(1)/(2)-(1)/(4)b \end{gathered}`

Replacing the value of 2a" in equation (8)

`35(-(1)/(2)-(1)/(4)b)+5b=5`

Solving the operations:

`-(35)/(2)-(35)/(4)b+5b=5`

Solving for "b":

`\begin{gathered} -(7)/(2)-(7)/(4)b+b=1 \\ -(7)/(2)-(3)/(4)b=1 \end{gathered}`

Adding 7/2 to both sides:

`\begin{gathered} -(3)/(4)b=1+(7)/(2) \\ -(3)/(4)b=(9)/(2) \\ b=-6 \end{gathered}`

Therefore, b = 9.

Replacing this value in equation (7)

`\begin{gathered} a=-(1)/(2)-(1)/(4)(-6) \\ a=1 \end{gathered}`

Replacing in equation (2)

`\begin{gathered} 1-6+c=-2 \\ \end{gathered}`

Adding like terms:

`-5+c=-2`

Adding 5 to both sides:

`\begin{gathered} c=-2+5 \\ c=3 \end{gathered}`

Therefore, the quadratic equation is:

`y=x^2-6x+3`