Question

use a system of equations to find the parabola of the form $y = ax^2 + bx + c$ that goes through the three given points.\
$(3, 12)$, $(-3, 30)$, $(-2, 17)$\
the parabola fitting these three points is $y = \square$.

Explanation:

Step1: Substitute points into equation

For point \((3, 12)\): \(12 = a(3)^2 + b(3) + c\) → \(9a + 3b + c = 12\)
For point \((-3, 30)\): \(30 = a(-3)^2 + b(-3) + c\) → \(9a - 3b + c = 30\)
For point \((-2, 17)\): \(17 = a(-2)^2 + b(-2) + c\) → \(4a - 2b + c = 17\)

Step2: Subtract first two equations

\((9a - 3b + c) - (9a + 3b + c) = 30 - 12\)
\(-6b = 18\) → \(b = -3\)

Step3: Substitute \(b = -3\) into first and third equations

First: \(9a + 3(-3) + c = 12\) → \(9a + c = 21\)
Third: \(4a - 2(-3) + c = 17\) → \(4a + c = 11\)

Step4: Subtract third from first (with \(b=-3\))

\((9a + c) - (4a + c) = 21 - 11\)
\(5a = 10\) → \(a = 2\)

Step5: Find \(c\) using \(a=2\) and \(9a + c = 21\)

\(9(2) + c = 21\) → \(18 + c = 21\) → \(c = 3\)

Answer:

\(2x^2 - 3x + 3\)