Calculus
posted by Cad on .
Find the equation of a quartic polynomial whose graph is symmetric about the yaxis and has local maxima at (−1,1) and (1,1) and a yintercept of 1.

y = a x^4 + b x^3 + c x^2 + d x + e
when x = 0, y = 1 so e = 1
so
y = a x^4 + b x^3 + c x^2 + d x  1
for max or min
dy/dx = 0 = 4 a x^3 + 3 b x^2 + +2 c x + d
when x = 1
0 = 4 a (1) +3b(1)+2c(1)+d
1 = a (1) + b(1) + c(1) +d(1)  1
when x = +1
0 = 4a + 3 b + 2c + d
1 = a+b+c+d1

start solving
0 = 4a+3b2c+d
0 = +4a+3b+2c+d

0 = 6b +2d
2 = ab+cd
2 = a+b+c+d

0 = 2b2d
0 = 6b +2d
0 = 2b2d

8 b = 0
b=0
then d = 0
(in fact from symmetry I bet all odd terms are 0)
now go back and get a and c