Posted by weloo_volley on Saturday, April 21, 2012 at 6:15pm.
y = a x^3 + b x^2 + c x + d
y' = 3 a x^2 + 2 b x + c
y" = 6 a x + 2 b
y" = 0 at x = 2 so 12 a + 2 b = 0
y' = 0 at x = -1 so 3a - 2b + c = 0
y = 8/3 at x = -1 so -a + b -c + d = 8/3
y =-46/3 at x=2 so 8a+ 4b+ 2c+ d = -46/3
----------------solve
b = -6 a
3a - 2b + c = 0 --> 15 a + c = 0
so c = -15 a
-a + b -c + d = 8/3
so
-a -6a +15 a + d = 8/3
8 a + d = 8/3 so d = 8/3 - 8a
then last one
8a+ 4b+ 2c+ d = -46/3
8a +4(-6a)+2(-15a)+8/3-8a = -46/3
a(8-24-30-8) = -54/3
-54 a = -54/3
a = 1/3
onward from there to b , c , d
the inflection point is halfway between critical points so another critical point at x = 5
y' = 0 at x = +5 so 75 a +10 b + c = 0
in summary
12 a + 2 b = 0 so b = -6 a
3a - 2b + c = 0
75 a + 10 b + c = 0
==========================
3 a -2(-6a) + c = 0
75 a +10(-6a) + c = 0
15 a + c = 0
15 a + c = 0
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