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- Homework Statement
- The curves y=x^2+ax+b and y=cx-x^2 have a common tangent line at the point (1,0). Find a, b, and c
- Relevant Equations
- Derivatives of both: 2x+a, and c-2x
I do know that since they have a common tangent line, that means:
2x+a = c-2x
Since they both have the point (1,0), then since both equations should equal 0 when x = 1:
c(1)-(1)^2 = 0 --> c = 1
So now, I replace c with 1 to solve for a in the two derivatives that are equal (common tangent line) at x = 1:
2(1)+a = 1-2(1) --> a = -3
Then last, I plug x and a values into the first equation at (1,0):
(1)^2+(-3)(1)+b = 0 --> b = 2
When I plug those values into the original equations for x = 1 and y = 0, they check out. When I set the derivatives equal to each other, that checks out (both sides = -1). I think I did this right, but for some reason, my solutions manual has entirely different solutions to entirely different problems for the last half of this chapter in Thomas' Calculus (even though they're both 13th ed), so I can't verify it for sure. Can someone please let me know if this is all correct?
2x+a = c-2x
Since they both have the point (1,0), then since both equations should equal 0 when x = 1:
c(1)-(1)^2 = 0 --> c = 1
So now, I replace c with 1 to solve for a in the two derivatives that are equal (common tangent line) at x = 1:
2(1)+a = 1-2(1) --> a = -3
Then last, I plug x and a values into the first equation at (1,0):
(1)^2+(-3)(1)+b = 0 --> b = 2
When I plug those values into the original equations for x = 1 and y = 0, they check out. When I set the derivatives equal to each other, that checks out (both sides = -1). I think I did this right, but for some reason, my solutions manual has entirely different solutions to entirely different problems for the last half of this chapter in Thomas' Calculus (even though they're both 13th ed), so I can't verify it for sure. Can someone please let me know if this is all correct?