Calculus Problem Solving Question

danizh
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Please help me solve this problem.

Find numbers a, b, and c so that the graph of f(x) = ax^2 + bx + c has x-intercepts at (0,0) and (8,0) and a tangent with slope 16 where x = 2.

I have done this so far:
f(x) = ax^2 + bx + c
f '(x) = (2)(a)(x) + b
16 = (2)(a)(2) + b
16 = 4a + b

I don't know where to go from here, so any help would be greatly appreciated.
 
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Nevermind, I solved it. If anyone else needs the solution to this problem, let me know. :)
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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