Question

use quadratic regression to find a function that fits the following points. (-2, -1) (1, -1) (3, 9) ?x² + x +

Explanation:

Step1: Write the general quadratic form

The general form of a quadratic function is $y = ax^{2}+bx + c$. Substitute the given points into this equation.
For the point $(-2,-1)$: $-1=a(-2)^{2}+b(-2)+c = 4a-2b + c$.
For the point $(1,-1)$: $-1=a(1)^{2}+b(1)+c=a + b + c$.
For the point $(3,9)$: $9=a(3)^{2}+b(3)+c = 9a+3b + c$.

Step2: Create a system of equations

We have the following system of equations:

$$\begin{cases} 4a-2b + c=-1\\ a + b + c=-1\\ 9a+3b + c=9 \end{cases}$$

Subtract the second - equation from the first equation:
$(4a-2b + c)-(a + b + c)=-1-(-1)$.
$4a-2b + c - a - b - c=-1 + 1$.
$3a-3b=0$, which simplifies to $a - b=0$ or $a = b$.
Subtract the second - equation from the third equation:
$(9a+3b + c)-(a + b + c)=9-(-1)$.
$9a+3b + c - a - b - c=9 + 1$.
$8a+2b=10$.

Step3: Substitute $a = b$ into $8a+2b=10$

Since $a = b$, then $8a+2a=10$.
$10a=10$, so $a = 1$ and $b = 1$.

Step4: Find the value of $c$

Substitute $a = 1$ and $b = 1$ into the equation $a + b + c=-1$.
$1+1 + c=-1$.
$c=-1-2=-3$.

Answer:

$1x^{2}+1x - 3$