Find real numbers \( a \), \( b \), and \( c \) so that the graph of the function \( y = ax^2 + bx + c \) contains the points \((-1, 6)\), \( (2, 7) \), and \( (0, 1) \).
Answer (2)
This gives you three simultaneous equations:
6 = a + c 7 = 4a + c 1 = c
c = 1 **** If c =1,
6 = a + 1 a = 5 **** This doesn't work in the second equation, so the quadratic that goes through these points is not in the form y = ax^2 + bx + c Was there supposed to be a b in the equation?
The values of the coefficients for the quadratic function that passes through the given points are a = 3 8 , b = − 3 7 , and c = 1 . Thus, the function is y = 3 8 x 2 − 3 7 x + 1 .
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