Question
Which parabola passes through the points (1, -2), (4, 1), and (5, -2)
Answers
GPT 3.5
To find the equation of the parabola passing through the given points, we can start by assuming that the general form of a parabola is y = ax^2 + bx + c.
We can substitute the coordinates of the first point, (1,-2), into the equation to get:
-2 = a(1^2) + b(1) + c
-2 = a + b + c ---(1)
Similarly, substituting the coordinates of the second point, (4,1), into the equation, we get:
1 = a(4^2) + b(4) + c
1 = 16a + 4b + c ---(2)
And substituting the coordinates of the third point, (5,-2), into the equation, we get:
-2 = a(5^2) + b(5) + c
-2 = 25a + 5b + c ---(3)
From equations (1), (2), and (3), we can form a system of linear equations:
a + b + c = -2 ---(1)
16a + 4b + c = 1 ---(2)
25a + 5b + c = -2 ---(3)
To solve this system of equations, we can use any method such as substitution or elimination. Let's use the elimination method to solve this system.
We can subtract equation (3) from equation (2) to eliminate c:
(16a + 4b + c) - (25a + 5b + c) = 1 - (-2)
16a + 4b - 25a - 5b = 1 + 2
-9a - b = 3 ---(4)
Subtracting equation (1) from equation (2) also eliminates c:
(16a + 4b + c) - (a + b + c) = 1 - (-2)
16a + 4b - a - b = 1 + 2
15a + 3b = 3 ---(5)
Now, we have a system of two equations:
-9a - b = 3 ---(4)
15a + 3b = 3 ---(5)
Multiplying equation (4) by 3 and equation (5) by 9 to eliminate b, we get:
-27a - 3b = 9 ---(6)
135a + 27b = 27 ---(7)
Adding equation (6) to equation (7) will eliminate b:
(-27a - 3b) + (135a + 27b) = 9 + 27
108a = 36
a = 36/108
a = 1/3
Substituting the value of a into equation (4), we can solve for b:
-9a - b = 3
-9(1/3) - b = 3
-3 - b = 3
b = -3 -3
b = -6
Substituting the values of a and b into equation (1), we can solve for c:
a + b + c = -2
(1/3) + (-6) + c = -2
(1/3) -6 + c = -2
c - 18 + 3c = -6
4c - 18 = -6
4c = 18 - 6
4c = 12
c = 12/4
c = 3
Therefore, the equation of the parabola passing through the points (1, -2), (4, 1), and (5, -2) is y = (1/3)x^2 - 6x + 3.
We can substitute the coordinates of the first point, (1,-2), into the equation to get:
-2 = a(1^2) + b(1) + c
-2 = a + b + c ---(1)
Similarly, substituting the coordinates of the second point, (4,1), into the equation, we get:
1 = a(4^2) + b(4) + c
1 = 16a + 4b + c ---(2)
And substituting the coordinates of the third point, (5,-2), into the equation, we get:
-2 = a(5^2) + b(5) + c
-2 = 25a + 5b + c ---(3)
From equations (1), (2), and (3), we can form a system of linear equations:
a + b + c = -2 ---(1)
16a + 4b + c = 1 ---(2)
25a + 5b + c = -2 ---(3)
To solve this system of equations, we can use any method such as substitution or elimination. Let's use the elimination method to solve this system.
We can subtract equation (3) from equation (2) to eliminate c:
(16a + 4b + c) - (25a + 5b + c) = 1 - (-2)
16a + 4b - 25a - 5b = 1 + 2
-9a - b = 3 ---(4)
Subtracting equation (1) from equation (2) also eliminates c:
(16a + 4b + c) - (a + b + c) = 1 - (-2)
16a + 4b - a - b = 1 + 2
15a + 3b = 3 ---(5)
Now, we have a system of two equations:
-9a - b = 3 ---(4)
15a + 3b = 3 ---(5)
Multiplying equation (4) by 3 and equation (5) by 9 to eliminate b, we get:
-27a - 3b = 9 ---(6)
135a + 27b = 27 ---(7)
Adding equation (6) to equation (7) will eliminate b:
(-27a - 3b) + (135a + 27b) = 9 + 27
108a = 36
a = 36/108
a = 1/3
Substituting the value of a into equation (4), we can solve for b:
-9a - b = 3
-9(1/3) - b = 3
-3 - b = 3
b = -3 -3
b = -6
Substituting the values of a and b into equation (1), we can solve for c:
a + b + c = -2
(1/3) + (-6) + c = -2
(1/3) -6 + c = -2
c - 18 + 3c = -6
4c - 18 = -6
4c = 18 - 6
4c = 12
c = 12/4
c = 3
Therefore, the equation of the parabola passing through the points (1, -2), (4, 1), and (5, -2) is y = (1/3)x^2 - 6x + 3.