To solve this quadratic regression problem, we need to find the equation of a parabola in the form:

$y = ax^2 + bx + c$

given the three points: $(5, 210)$, $(1, 14)$, and $(-3, 10)$.

Step-by-Step Solution

1. Set up the system of equations: For each point $(x, y)$, substitute the values of $x$ and $y$ into the equation $y = ax^2 + bx + c$.

   • For $(5, 210)$: $210 = 25a + 5b + c$
   • For $(1, 14)$: $14 = a + b + c$
   • For $(-3, 10)$: $10 = 9a - 3b + c$

2. Write the system of linear equations:

   25a + 5b + c = 210 \\ a + b + c = 14 \\ 9a - 3b + c = 10 \end{cases}$$

3. Solve the system of equations: We can solve these equations by elimination or substitution to find the values of $a$, $b$, and $c$.

After solving this system, you’ll find the values of $a$, $b$, and $c$ that define the quadratic function. Let me calculate these values for you.The solution to the system of equations gives us the values:

$a = 6, \quad b = 13, \quad c = -5$

Thus, the equation of the parabola passing through the points $(5, 210)$, $(1, 14)$, and $(-3, 10)$ is:

$y = 6x^2 + 13x - 5$

Would you like a breakdown of the steps used to solve the system, or do you have any other questions?

Here are some related questions that might help expand your understanding:

1. How would the process change if the points did not lie on a parabola?
2. What other methods can be used for finding the best-fit quadratic equation if the points only approximately fit?
3. How can you verify if a quadratic equation correctly passes through given points?
4. What applications in real life require finding a quadratic equation that passes through certain points?
5. How does the quadratic regression method compare to linear regression?

Tip: When solving systems of equations with three variables, substitution or elimination methods can be used effectively, and tools like matrices or computational software can simplify complex calculations.