Question 703960: 16. Find a quadratic function in standard form that has the following points: (1, -2); (2, -2); (3, -4).
Answer by DrBeeee(684)   (Show Source):
You can put this solution on YOUR website! Use the standard form
(1) y = ax^2 + bx + c
Substitute (1,-2) into (1) and get
(2) -2 = a + b + c
Substitute (2,-2) into (1) and get
(3) -2 = 4a + 2b + c
Substitute (3,-4) into (1) and get
(4) -4 = 9a + 3b + c
We have three equations and three unknowns. All we need to do is solve for a, b, and c.
Subtract (3) from (4) and get
(5) -2 = 5a + b
Subtract (2) from (3) and get
(6) 0 = 3a + b
Now subtract (6) from (5) and get
(7) -2 = 2a or
(8) a = -1
Put (8) into (6) and get
(9) b = 3
Put a and b into (2) and get
(10) c = -4
Let's check the equation with the three points.
Is (-2 = -1*1 + 3*1 -4)?
Is (-2 = -1 + 3 - 4)?
Is (-2 = -2)? Yes
Is (-2 = -1*2^2 + 3*2 - 4)?
Is (-2 = -4 +6 -4)?
Is (-2 = -2)? Yes
Is (-4 = -1*3^2 + 3*3 -4)?
Is (-4 = -9 +9 -4)?
Is (-4 = -4)? Yes
Answer: The quadratic function (parabola) that passes through the three given points is -x^2 + 3x -4. By the way, the parabola's apex is (3/2,-7/4)
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