use quadratic regression to find a function that fits the following points. (1, -8) (2, -4) (3, 6) ?x² + x +

Explanation:

Step1: Set up the quadratic equation

Let the quadratic function be $y = ax^{2}+bx + c$. Substitute the given points $(1,-8)$, $(2,-4)$ and $(3,6)$ into the equation. For $(1,-8)$: $a\times1^{2}+b\times1 + c=-8$, which simplifies to $a + b + c=-8$. For $(2,-4)$: $a\times2^{2}+b\times2 + c=-4$, which simplifies to $4a+2b + c=-4$. For $(3,6)$: $a\times3^{2}+b\times3 + c=6$, which simplifies to $9a+3b + c=6$.

Step2: Create a system of linear - equations

We have the following system of equations: $\begin{cases}a + b + c=-8\4a+2b + c=-4\9a+3b + c=6\end{cases}$ Subtract the first equation from the second equation: $(4a + 2b + c)-(a + b + c)=-4-(-8)$ $4a+2b + c - a - b - c=-4 + 8$ $3a + b=4$. Subtract the second equation from the third equation: $(9a+3b + c)-(4a + 2b + c)=6-(-4)$ $9a+3b + c - 4a - 2b - c=6 + 4$ $5a + b=10$.

Step3: Solve the new system of equations

We have the system $\begin{cases}3a + b=4\5a + b=10\end{cases}$ Subtract the first new - equation from the second new - equation: $(5a + b)-(3a + b)=10 - 4$ $5a + b-3a - b=6$ $2a=6$, so $a = 3$. Substitute $a = 3$ into $3a + b=4$: $3\times3 + b=4$ $9 + b=4$, so $b=-5$.

Step4: Find the value of c

Substitute $a = 3$ and $b=-5$ into $a + b + c=-8$: $3-5 + c=-8$ $-2 + c=-8$, so $c=-6$.

Answer:

$3x^{2}-5x - 6$