danizh
Oct6-05, 10:26 PM
Please help me solve this problem. :yuck:
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.
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.