Find 2nd-degree polynomial intersecting 4th-degree polynomial
- Thread starter swhylina
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Please show us what you have tried and exactly where you are stuck.
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Dr.Peterson
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Have you looked for a polynomial based on (ii) that fits the description of (i)?
Otis
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We hope that swhylina was able to finish the exercise. 
14c(i) [imath]\;[/imath] P(x) = (x–m)^2 * (x–n)^2
14c(ii) [imath]\;[/imath] a = 15/28 [imath]\;[/imath] and [imath]\;[/imath] b = 64/49
I took advantage of the instruction "or otherwise" and used first derivatives of f and g to write a system of four equations. Let the tangent points occur at x=c (Quadrant II) and x=d (Quadrant I).
f(c) = g(c)
f'(c) = g'(c)
f(d) = g(d)
f'(d) = g'(d)
Edit: Before using derivatives, I'd estimated slopes (by guessing values for c and d) using the standard formula. A pair of numerical processes gave me a decent estimate for parameter a (which showed a repeating decimal pattern), allowing me to guess its actual value without formal calculus. The entire approach involved too much rigamarole to post.
[imath]\;[/imath]
14c(i) [imath]\;[/imath] P(x) = (x–m)^2 * (x–n)^2
14c(ii) [imath]\;[/imath] a = 15/28 [imath]\;[/imath] and [imath]\;[/imath] b = 64/49
I took advantage of the instruction "or otherwise" and used first derivatives of f and g to write a system of four equations. Let the tangent points occur at x=c (Quadrant II) and x=d (Quadrant I).
f(c) = g(c)
f'(c) = g'(c)
f(d) = g(d)
f'(d) = g'(d)
Edit: Before using derivatives, I'd estimated slopes (by guessing values for c and d) using the standard formula. A pair of numerical processes gave me a decent estimate for parameter a (which showed a repeating decimal pattern), allowing me to guess its actual value without formal calculus. The entire approach involved too much rigamarole to post.
[imath]\;[/imath]