# Calculus

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Find the equation of a quartic polynomial whose graph is symmetric about the y-axis and has local maxima at (−1,1) and (1,1) and a y-intercept of -1.

• Calculus -

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+d-1
--------------------------------
start solving
0 = -4a+3b-2c+d
0 = +4a+3b+2c+d
-----------------
0 = 6b +2d

2 = a-b+c-d
2 = a+b+c+d
-------------
0 = -2b-2d

0 = 6b +2d
0 = -2b-2d
-----------
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

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