Question 857380: The quadratic function which takes the value 41 at x=-2 and the value 20 at x=5 and is minimized at x=2 is: y=Ax^2 - Bx + C, The minimum value of this function is D. (please solve for ABC and D, thank you)
Answer by richwmiller(17219)   (Show Source):
You can put this solution on YOUR website! 1) plug in -2 for x and 41 for y
41=A(-2)^2 - B(-2) + C
41=4A+4B+C
2) plug in 5 for x and 20 for y
20=A(5)^2 - 5B + C
20=25A-5B+C
41=4A+4B+C
-21=21A-9B
vertex x=2=-B/2A
3) Subtract and eliminate C
-21=21A-9B
2=-B/2A
solve for A & B
A = -7/19, B = 28/19, C = 695/19
4) plug in A and B and solve for C
plug in 2 for x and solve for D
A = -7/19, B = 28/19, C = 695/19, D= 611/19
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