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- Anthony Chun
- For Ned: Calculus 3/1/2015
- f(x) = Ax^3 + Bx^2 + Cx + D
- The tangent line to it at x = 0 is y = x
- The tangent line at x = 2 is y = 2x - 3
- When we plug x = 0 to y = x, we get y = 0.
- This is one point: (0,0)
- When we plug in x = 2 to y = 2x - 3, we get y = 2(2) - 3 = 1
- This is another point, (2,1)
- Ok so tangent lines involve the derivative.
- The derivative becomes the coefficient of x for the tangent line.
- Coefficient of the first one is 1, and the second one is 2.
- So F'(x) = 3Ax^2 + 2Bx + C
- Now that we have all of this analyzed, we can solve the problem.
- We know that one point on the graph is (0,0).
- We can plug this in to the first equation f(x).
- 0 = A(0)^3 + B(0)^2 + C(0) + D
- 0 = D
- We also know from y = x being the tangent line, the slope(derivative) at 0 is 1.
- 1 = 3A(0)^2 + 2B(0) + C
- This means C = 1
- We now have two variables already, C = 1, D = 0.
- Let's continue.
- Now we have specific points that need to be on the graph, lets plug them into f(x) and f'(x).
- We know that the slope(derivative) at x = 2 is 2.
- We plug (2,2) into f'(x).
- 2 = 3A(2)^2 + 2B(2) + 1
- 1 = 12A + 4B
- We also have another point on the graph: (2,1). We plug it into f(x).
- 1 = A(2)^3 + B(2)^2 + 1(2) + 0
- 1 = 8A + 4B + 2
- -1 = 8A + 4B
- With the above two we have
- 1 = 12A + 4B
- -1 = 8A + 4B
- Subtract the two equations.
- 2 = 4A
- A = 1/2
- Plug it back into one of the equations
- -1 = 8(1/2) + 4B
- -1 = 4 + 4B
- -5 = 4B
- B = -5/4
- We have
- A = 1/2
- B = -5/4
- C = 1
- D = 0
- Our final equation is
- y = (1/2)x^3 - (5/4)x^2 + x
- Here are the graphs from x = -2 to 4
- http://aton2.com/cap/8601bcbd1a9e9f751902.png
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