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Question
Find A, B, C, D such that the graph of f(x)= Ax^3 + Bx^2 + Cx + D is tangent to the line y = 3x -3 at the point (1, 0) and is tangent to the line y = 18x -27 at the point(2, 9)
Attir~
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This is not an easy problem since solving the system gets messy algebraically but the concepts are easy. You are given f(x) where there are two point on f(x) namely (1,0) and (2,9). Since they are tangent to y = 3x -3 at (1,0) and y = 18x -27 at (2,9) these are the point of intersection also. You therefore has f(1)= A + B + C + D = 0 and
f(2) = 8A + 4B + 2C + D = 9 as two equations in your system to solve. take the first derivative f'(x) = 3Ax^2 + 2Bx + C and plug in the points there as well since the first derivative at each of the points is the equation of the tangent line which will be each of y = 3x -3 and y = 18x -27. Attached you will find a complete explanation. I typed it in and before I was done it disappeared so I hand wrote it :).
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Math Prof
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Sherry Wallin
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I can answer most questions up through Calculus and some in Number Theory and Abstract Algebra.
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I have had my Bachelor's Degree since 1987 and have been a teacher since 1988. I earned my Masters Degree in Mathematics May 2010. I have been teaching at the same community college since 2002.
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