The problem asks to use quadratic regression to find the equation of the parabola passing through the points (-26, -22), (-10, 154), and (3, -80). The quadratic equation has the form:

$y = ax^2 + bx + c$

We will substitute the given points into this general equation to create a system of three equations, then solve for the unknowns $a$, $b$, and $c$.

Step 1: Write equations for the points

For (-26, -22): $-22 = a(-26)^2 + b(-26) + c \implies -22 = 676a - 26b + c$

For (-10, 154): $154 = a(-10)^2 + b(-10) + c \implies 154 = 100a - 10b + c$

For (3, -80): $-80 = a(3)^2 + b(3) + c \implies -80 = 9a + 3b + c$

Step 2: Solve the system of equations

We now have three linear equations:

1. $676a - 26b + c = -22$
2. $100a - 10b + c = 154$
3. $9a + 3b + c = -80$

I will solve this system for $a$, $b$, and $c$.The equation of the parabola passing through the points $(-26, -22)$, $(-10, 154)$, and $(3, -80)$ is:

$y = -x^2 - 25x + 4$

Let me know if you'd like further explanation or have any questions.

Here are 5 related questions:

1. How do you solve a system of three equations in general?
2. What is quadratic regression, and how does it differ from linear regression?
3. How can you check if the equation obtained is correct?
4. How would the equation change if a fourth point was added?
5. What are the practical applications of quadratic regression in real life?

Tip: When solving quadratic equations for specific points, always ensure the equation fits all points by substituting the values back into the equation.