Question 190090: Find the equation of the quadratic function with vertex (6, -40) and passing through the point (3,50)
Answer by stanbon(75887) (Show Source):
You can put this solution on YOUR website! Find the equation of the quadratic function with vertex (6, -40) and passing through the point (3,50)
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The vertex occurs when x = -b/2a = 6
So 12a = -b
b = -12a
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General form of quadratic:
y = ax^2 + bx + c
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Substitute to get:
y = ax^2 -12ax + c
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Since the function passes thru (3,50), substitute to solve for "c".
50 = a(9) -12a(3) + c
50 = -27a + c
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Since the function passes thru (6.-40)
-40 = a(36) -12a(6) + c
-40 = -36a + c
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Solve 2-equation system by substitution:
c = 27a+50
c = 36a-40
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27a+50 = 36a-40
9a = 90
a = 10
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Solve for "c":
c = 36(10)-40 = 320
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Since b = -12a, b = -120
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Equation: y = 10x^2 -120x + 320
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Cheers,
Stan H.
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