Answer
$y=-\dfrac{5}{6}x^2+\dfrac{13}{2}x-\dfrac{35}{3}$
Work Step by Step
We have to determine $a,b,c$ so that the graph of the function $y=ax^2+bx+c$ passes through the points $(1,-6),(2,-2),(4,1)$.
Use the fact that each of the given points $(x,y)$ satisfies the equation $y=ax^2+bx+x$.
We find the system:
$\begin{cases}
a(1)^2+b(1)+c=-6\\
a(2)^2+b(2)+c=-2\\
a(4)^2+b(4)+c=1
\end{cases}$
$\begin{cases}
a+b+c=-6\\
4a+2b+c=-2\\
16a+4b+c=1
\end{cases}$
We will use the addition method. Multiply Equation 1 by -1 and add it to Equation 2 and Equation 3 to eliminate $c$:
$\begin{cases}
4a+2b+c-a-b-c=-2+6\\
16a+4b+c-a-b-c=1+6
\end{cases}$
$\begin{cases}
3a+b=4\\
15a+3b=7
\end{cases}$
Multiply Equation 1 by -3 and add it to Equation 2 to eliminate $b$ and determine $a$:
$\begin{cases}
-3(3a+b)=-3(4)\\
15a+3b=7
\end{cases}$
$\begin{cases}
-9a-3b=-12\\
15a+3b=7
\end{cases}$
$-9a-3b+15a+3b=-12+7$
$6a=-5$
$a=-\dfrac{5}{6}$
Determine $b$ using the equation $3a+b=4$:
$3\left(-\dfrac{5}{6}\right)+b=4$
$-\dfrac{5}{2}+b=4$
$b=4+\dfrac{5}{2}$
$b=\dfrac{13}{2}$
Determine $c$ by substituting $a,b$ in Equation 1:
$a+b+c=-6$
$-\dfrac{5}{6}+\dfrac{13}{2}+c=-6$
$c=-6-\dfrac{34}{6}$
$c=-\dfrac{35}{3}$
The system's solution is:
$\left(-\dfrac{5}{6},\dfrac{13}{2},-\dfrac{35}{3}\right)$
The function is fully determined:
$y=-\dfrac{5}{6}x^2+\dfrac{13}{2}x-\dfrac{35}{3}$
