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):

$\begin{gathered} -8a=4+2b \\ a=-\frac{1}{2}-\frac{1}{4}b \end{gathered}$

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

$35(-\frac{1}{2}-\frac{1}{4}b)+5b=5$

Solving the operations:

$-\frac{35}{2}-\frac{35}{4}b+5b=5$

Solving for "b":

$\begin{gathered} -\frac{7}{2}-\frac{7}{4}b+b=1 \\ -\frac{7}{2}-\frac{3}{4}b=1 \end{gathered}$

Adding 7/2 to both sides:

$\begin{gathered} -\frac{3}{4}b=1+\frac{7}{2} \\ -\frac{3}{4}b=\frac{9}{2} \\ b=-6 \end{gathered}$

Therefore, b = 9.

Replacing this value in equation (7)

$\begin{gathered} a=-\frac{1}{2}-\frac{1}{4}(-6) \\ a=1 \end{gathered}$

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:

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