If you want y = e^(-(x-a)^2/b) then
y' = -1/b (2(x-a)) e^(-(x-a)^2/b)
y" = [-1/b (2) + -1/b (2(x-a)) [-1/b (2(x-a))]] e^(-(x-a)^2/b)
= (4(x-a)^2/b^2 - 2/b) e^(-(x-a)^2/b)
= 2/b^2 (2(x-a)^2 - b) e^(-(x-a)^2/b)
So, y"=0 at x=a±√(b/2)
3 = 6-3
9 = 6+3
so we need a=6, b=12
That gives us
y = e^(-(x-6)^2/12)
see the plot at
wolframalpha. com/input?i=+e%5E%28-%28x-6%29%5E2%2F12%29
Find a formula for a curve of the form y=e−(x−a)2/b
for b>0 with a local maximum at x=6 and points of inflection at x=3 and x=9.
1 answer