use quadratic regression to find a function that fits the following points. (-2, -16) (3, 11) (0, 2) ?x² + x…

Explanation:

Step1: Set up the quadratic function

Let the quadratic function be $y = ax^{2}+bx + c$. Substitute the given points into the function. For point $(-2,-16)$: $-16=a(-2)^{2}+b(-2)+c$, which simplifies to $4a - 2b + c=-16$. For point $(3,11)$: $11=a(3)^{2}+b(3)+c$, which simplifies to $9a+3b + c = 11$. For point $(0,2)$: $2=a(0)^{2}+b(0)+c$, so $c = 2$.

Step2: Substitute $c = 2$ into the other two - equations

Substitute $c = 2$ into $4a - 2b + c=-16$ and $9a+3b + c = 11$. We get $4a-2b+2=-16$, which simplifies to $4a - 2b=-18$ (divide by 2 to get $2a - b=-9$). And $9a+3b+2 = 11$, which simplifies to $9a+3b = 9$ (divide by 3 to get $3a + b = 3$).

Step3: Solve the system of equations

Add the equations $2a - b=-9$ and $3a + b = 3$ together. $(2a - b)+(3a + b)=-9 + 3$. $2a - b+3a + b=-6$. $5a=-6$, so $a=-\frac{6}{5}=-1.2$.

Step4: Find the value of $b$

Substitute $a=-1.2$ into $3a + b = 3$. $3(-1.2)+b = 3$. $-3.6 + b = 3$. $b=3 + 3.6=6.6$.

Answer:

$-1.2x^{2}+6.6x + 2$