Question 1062481: determine a possible quadratic function for points (2,7), (3,5) and (5,13).
Found 2 solutions by josgarithmetic, Alan3354:
Answer by josgarithmetic(39617)   (Show Source):
You can put this solution on YOUR website! You have a choice.
x=ay^2+by+c if you want symmetry axis parallel to the x-axis;
y=ax^2+bx+c if you want symmetry axis parallel to the y-axis;
or something with xy term if some other symmetry axis.
Each given point gives a separate, specific equation. Make THREE such equations, and solve for a, b, c, in the linear system. Do you need to see this done or started?
Answer by Alan3354(69443)  (Show Source):
You can put this solution on YOUR website! determine a possible quadratic function for points (2,7), (3,5) and (5,13).
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2 parabolas can be found that pass thru the points.
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y = ax^2 + bx + c is one.
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4a + 2b + c = 7
9a + 3b + c = 5
25a + 5b + c = 13
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Eqn2 - Eqn 1:
5a + b = -2 **** Eqn 4
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Eqn 3 - Eqn 2:
16a + 2b = 8
10a + 2b = -4 Eqn 4 times 2
---------------------------------- Subtract
6a = 12
a = 2
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b = -12
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c = 23
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f(x) = 2x^2 - 12x + 23
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Find the 2nd one the same way, y as the independent variable.
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